删除重复项和对向量进行排序的最有效方法是什么?


274

我需要使用可能包含很多元素的C ++向量,删除重复项并对其进行排序。

我目前有以下代码,但无法正常工作。

vec.erase(
      std::unique(vec.begin(), vec.end()),
      vec.end());
std::sort(vec.begin(), vec.end());

如何正确执行此操作?

另外,先擦除重复项(类似于上面的代码)还是先执行排序更快?如果我确实先执行排序,是否保证std::unique执行后仍保持排序?

还是有另一种(也许更有效)的方式来完成所有这些工作?


3
我假设您没有选择插入之前进行检查的选项,以避免一开始就出现重复?

正确。那将是理想的。
Kyle Ryan

29
我建议您更正上面的代码,或者真正表明它是错误的。std :: unique假定范围已排序。
Matthieu M.

Answers:


584

我同意R. PateTodd Gardner的观点 ; 一个std::set可能会在这里是个好主意。即使您坚持使用向量,但如果有足够的重复项,则最好创建一个集合来完成肮脏的工作。

让我们比较三种方法:

仅使用向量,排序+唯一

sort( vec.begin(), vec.end() );
vec.erase( unique( vec.begin(), vec.end() ), vec.end() );

转换为设置(手动)

set<int> s;
unsigned size = vec.size();
for( unsigned i = 0; i < size; ++i ) s.insert( vec[i] );
vec.assign( s.begin(), s.end() );

转换为集合(使用构造函数)

set<int> s( vec.begin(), vec.end() );
vec.assign( s.begin(), s.end() );

当重复数量发生变化时,它们的表现如下:

向量和集合方法的比较

简介:当重复项的数量足够大时,实际上转换为集合然后将数据转储为向量的速度更快

出于某种原因,手动执行集合转换似乎比使用集合构造函数更快-至少在我使用的玩具随机数据上。


61
我感到震惊的是,构造方法始终比手工方法差很多。您会发现除了一些微小的持续开销之外,它只会做手动的事情。谁能解释一下?
阿里

17
太好了,谢谢您的图表。您能否大致了解“重复次数”的单位是多少?(即“足够大”大约有多大?)?
Kyle Ryan

5
@Kyle:很大。我为该图使用了1000000个随机抽取的整数(在1到1000、100和10之间)的数据集。
Nate Kohl

5
我认为您的结果是错误的。在我的测试中,重复的元素越多,向量(比较)越快,实际上反方向缩放。您是否启用了优化编译功能并关闭了运行时检查功能?在我这边,向量总是更快,最高可达100倍,具体取决于重复项的数量。VS2013,cl / Ox -D_SECURE_SCL = 0。
davidnr 2014年

39
x轴的描述似乎丢失了。
BartoszKP '16

72

我重述了Nate Kohl的个人资料,并得到了不同的结果。对于我的测试用例,直接对向量进行排序总是比使用集合更有效。我使用新增了一种更有效的新方法unordered_set

请记住,unordered_set只有对于需要唯一和排序的类型具有良好哈希函数的方法,该方法才有效。对于整数,这很容易!(标准库提供了一个默认的哈希值,它只是标识函数。)而且,不要忘了在末尾进行排序,因为unordered_set是无序的:)

我在setunordered_set实现中进行了一些挖掘,发现构造函数实际上在为每个元素构造一个新节点之前,先检查其值以确定是否应将其实际插入(至少在Visual Studio实现中)。

这是5种方法:

f1:仅使用vectorsort+unique

sort( vec.begin(), vec.end() );
vec.erase( unique( vec.begin(), vec.end() ), vec.end() );

f2:转换为set(使用构造函数)

set<int> s( vec.begin(), vec.end() );
vec.assign( s.begin(), s.end() );

f3:转换为set(手动)

set<int> s;
for (int i : vec)
    s.insert(i);
vec.assign( s.begin(), s.end() );

f4:转换为unordered_set(使用构造函数)

unordered_set<int> s( vec.begin(), vec.end() );
vec.assign( s.begin(), s.end() );
sort( vec.begin(), vec.end() );

f5:转换为unordered_set(手动)

unordered_set<int> s;
for (int i : vec)
    s.insert(i);
vec.assign( s.begin(), s.end() );
sort( vec.begin(), vec.end() );

我使用100,000,000个整数(在[1,10],[1,1000]和[1,100000]范围内随机选择)进行了测试

结果(以秒为单位,越小越好):

range         f1       f2       f3       f4      f5
[1,10]      1.6821   7.6804   2.8232   6.2634  0.7980
[1,1000]    5.0773  13.3658   8.2235   7.6884  1.9861
[1,100000]  8.7955  32.1148  26.5485  13.3278  3.9822

4
对于整数,可以使用基数排序,这比std :: sort快得多。
Changming Sun

2
快速提示,使用sortunique方法,您必须#include <algorithm>
Davmrtl

3
@ChangmingSun我想知道为什么优化器似乎在f4上失败了?这些数字与f5截然不同。这对我来说没有任何意义。
桑索德18'Sep

1
@sandthorn如我的回答所述,该实现为输入序列中的每个元素构建一个节点(包括动态分配),这对于最终成为重复值的每个值都是浪费的。优化器不可能知道它可以跳过它。
alexk7

啊,这让我想起斯科特·迈耶的有关谈判一我CWUK 之情况有propablities的性质为减缓emplace这种结构。
桑索德18'Sep

57

std::unique 仅在重复元素相邻时才删除它们:您必须先对向量进行排序,然后向量才能按预期工作。

std::unique 被定义为稳定的,因此向量在其上运行唯一后仍将被排序。


42

我不确定您使用它的目的是什么,所以我不能100%肯定地说出来,但是通常当我认为“排序的,唯一的”容器时,我会想到std :: set。它可能更适合您的用例:

std::set<Foo> foos(vec.begin(), vec.end()); // both sorted & unique already

否则,在调用唯一之前进行排序(如其他答案所指出的)是可行的方法。


好吧!std :: set被指定为排序的唯一集。大多数实现使用高效的有序二叉树或类似的东西。
notnoop

+1想法也是如此。不想重复此答案
汤姆(Tom)在2009年

是否可以保证std :: set被排序?在实践中会是合理的,但是该标准是否要求?
MadCoder

1
是的,请参见23.1.4.9“关联容器的迭代器的基本属性是,它们以键的非降序在容器中进行迭代,其中键的降序由用于构造它们的比较来定义。”
Todd Gardner

1
@MadCoder:以某种排序的方式实现集合不一定“有意义”。还有一些使用哈希表实现的集,这些集未排序。实际上,大多数人喜欢在可用时使用哈希表。但是C ++中的命名约定恰好发生,以至于将排序后的关联容器简单地命名为“ set” /“ map”(类似于Java中的TreeSet / TreeMap)。和标准之外的散列关联容器称为“ hash_set” /“ hash_map”(SGI STL)或“ unordered_set” /“ unordered_map”(TR1)(类似于Java中的HashSet和HashMap)
newacct

22

std::unique仅适用于重复元素的连续运行,因此最好先进行排序。但是,它是稳定的,因此您的向量将保持排序。


18

这是为您执行此操作的模板:

template<typename T>
void removeDuplicates(std::vector<T>& vec)
{
    std::sort(vec.begin(), vec.end());
    vec.erase(std::unique(vec.begin(), vec.end()), vec.end());
}

像这样称呼它:

removeDuplicates<int>(vectorname);

2
+1模板化!-但你可以只写removeDuplicates(VEC),而不显式指定模板参数
费萨尔瓦利

10
甚至更好的方法是,直接使用模板化迭代器(开始和结束),并且可以在向量之外的其他结构上运行它。
凯尔·瑞安

地狱,模板!快速修复小清单,完整的STL样式。+1 thx
QuantumKarl 2015年

@Kyle-仅在具有erase()方法的其他容器上,否则您必须返回新的结束迭代器,并使调用代码截断该容器。
Toby Speight,2016年

8

效率是一个复杂的概念。需要考虑时间与空间的关系,以及常规度量(只能得到模糊的答案,例如O(n))与特定度量(例如,根据输入特性,气泡排序比快速排序要快得多)。

如果重复项相对较少,则排序方式是唯一的,然后进行擦除似乎是可行的方法。如果您有相对多的重复项,则从向量创建一个集合并让它繁重地工作很容易击败它。

也不要只专注于时间效率。Sort + unique + erase在O(1)空间中操作,而set构造在O(n)空间中操作。两者都没有直接适合于地图简化并行化(对于真正的大型数据集)。


什么会给您映射/缩小功能?我能想到的唯一一个是分布式合并排序,您仍然只能在最终合并中使用一个线程。
Zan Lynx 2009年

1
是的,您必须有一个控制节点/线程。但是,您可以根据需要对问题进行多次划分,以对控制/父线程处理的工作线程/子线程的数量以及每个叶节点必须处理的数据集的大小设置上限。并不是所有的问题都可以通过map-reduce轻松解决,我只是想指出,有些人处理类似的(无论如何,从表面上)优化问题,其中处理10 TB数据的过程称为“星期二”。

7

您需要在对其进行排序之前对其进行排序,unique因为这unique只会删除彼此相邻的重复项。

编辑:38秒...


7

unique只删除连续的重复元素(这对于线性运行是必需的),因此您应该首先执行排序。呼叫后,它将保持排序unique


7

如果您不想更改元素的顺序,则可以尝试以下解决方案:

template <class T>
void RemoveDuplicatesInVector(std::vector<T> & vec)
{
    set<T> values;
    vec.erase(std::remove_if(vec.begin(), vec.end(), [&](const T & value) { return !values.insert(value).second; }), vec.end());
}

也许使用unordered_set而不是set(和boost :: remove_erase_if如果可用)
gast128

4

假设a是向量,请使用删除连续的重复项

a.erase(unique(a.begin(),a.end()),a.end());运行时间为O(n)


1
连续的重复项。好的,所以需要std::sort先。
v.oddou

2

如前所述,unique需要一个分类的容器。此外,unique实际上并不从容器中删除元素。而是将它们复制到末尾,unique返回指向第一个此类重复元素的迭代器,并且您应该调用erase来实际删除这些元素。


唯一性是否需要排序的容器,还是仅重新排列输入序列,使其不包含相邻重复项?我以为是后者。

@Pate,您是对的。它不需要一个。它删除相邻的重复项。
比尔·林奇

如果您有一个可能包含重复项的容器,并且希望该容器在容器中的任何地方都没有任何重复的值,那么您必须首先对该容器进行排序,然后将其传递给唯一的容器,然后使用擦除操作来实际删除重复项。如果您只想删除相邻的重复项,则无需对容器进行排序。但是您最终会得到重复的值:如果将1 2 2 3 2 4 2 5 2传递给唯一而不进行排序,则将其更改为1 2 3 2 4 2 5 2;如果将其排序,则将1 2 3 4 5传递给唯一并擦除。 。
Max Lybbert 09年

2

Nate Kohl建议的标准方法,只是使用矢量,排序+唯一:

sort( vec.begin(), vec.end() );
vec.erase( unique( vec.begin(), vec.end() ), vec.end() );

不适用于指针向量。

在cplusplus.com上仔细查看此示例

在他们的示例中,移至末尾的“所谓重复项”实际上显示为?。(未定义的值),因为这些“所谓的重复项”是SOMETIMES的“额外元素”,而SOMETIMES则是原始向量中的“缺失元素”。

使用时发生问题 std::unique()指向对象的指针的向量(内存泄漏,从HEAP读取数据不正确,重复释放,导致分段错误等)。

这是我对问题的解决方案:替换std::unique()ptgi::unique()

请参阅下面的文件ptgi_unique.hpp:

// ptgi::unique()
//
// Fix a problem in std::unique(), such that none of the original elts in the collection are lost or duplicate.
// ptgi::unique() has the same interface as std::unique()
//
// There is the 2 argument version which calls the default operator== to compare elements.
//
// There is the 3 argument version, which you can pass a user defined functor for specialized comparison.
//
// ptgi::unique() is an improved version of std::unique() which doesn't looose any of the original data
// in the collection, nor does it create duplicates.
//
// After ptgi::unique(), every old element in the original collection is still present in the re-ordered collection,
// except that duplicates have been moved to a contiguous range [dupPosition, last) at the end.
//
// Thus on output:
//  [begin, dupPosition) range are unique elements.
//  [dupPosition, last) range are duplicates which can be removed.
// where:
//  [] means inclusive, and
//  () means exclusive.
//
// In the original std::unique() non-duplicates at end are moved downward toward beginning.
// In the improved ptgi:unique(), non-duplicates at end are swapped with duplicates near beginning.
//
// In addition if you have a collection of ptrs to objects, the regular std::unique() will loose memory,
// and can possibly delete the same pointer multiple times (leading to SEGMENTATION VIOLATION on Linux machines)
// but ptgi::unique() won't.  Use valgrind(1) to find such memory leak problems!!!
//
// NOTE: IF you have a vector of pointers, that is, std::vector<Object*>, then upon return from ptgi::unique()
// you would normally do the following to get rid of the duplicate objects in the HEAP:
//
//  // delete objects from HEAP
//  std::vector<Object*> objects;
//  for (iter = dupPosition; iter != objects.end(); ++iter)
//  {
//      delete (*iter);
//  }
//
//  // shrink the vector. But Object * pointers are NOT followed for duplicate deletes, this shrinks the vector.size())
//  objects.erase(dupPosition, objects.end));
//
// NOTE: But if you have a vector of objects, that is: std::vector<Object>, then upon return from ptgi::unique(), it
// suffices to just call vector:erase(, as erase will automatically call delete on each object in the
// [dupPosition, end) range for you:
//
//  std::vector<Object> objects;
//  objects.erase(dupPosition, last);
//
//==========================================================================================================
// Example of differences between std::unique() vs ptgi::unique().
//
//  Given:
//      int data[] = {10, 11, 21};
//
//  Given this functor: ArrayOfIntegersEqualByTen:
//      A functor which compares two integers a[i] and a[j] in an int a[] array, after division by 10:
//  
//  // given an int data[] array, remove consecutive duplicates from it.
//  // functor used for std::unique (BUGGY) or ptgi::unique(IMPROVED)
//
//  // Two numbers equal if, when divided by 10 (integer division), the quotients are the same.
//  // Hence 50..59 are equal, 60..69 are equal, etc.
//  struct ArrayOfIntegersEqualByTen: public std::equal_to<int>
//  {
//      bool operator() (const int& arg1, const int& arg2) const
//      {
//          return ((arg1/10) == (arg2/10));
//      }
//  };
//  
//  Now, if we call (problematic) std::unique( data, data+3, ArrayOfIntegersEqualByTen() );
//  
//  TEST1: BEFORE UNIQ: 10,11,21
//  TEST1: AFTER UNIQ: 10,21,21
//  DUP_INX=2
//  
//      PROBLEM: 11 is lost, and extra 21 has been added.
//  
//  More complicated example:
//  
//  TEST2: BEFORE UNIQ: 10,20,21,22,30,31,23,24,11
//  TEST2: AFTER UNIQ: 10,20,30,23,11,31,23,24,11
//  DUP_INX=5
//  
//      Problem: 21 and 22 are deleted.
//      Problem: 11 and 23 are duplicated.
//  
//  
//  NOW if ptgi::unique is called instead of std::unique, both problems go away:
//  
//  DEBUG: TEST1: NEW_WAY=1
//  TEST1: BEFORE UNIQ: 10,11,21
//  TEST1: AFTER UNIQ: 10,21,11
//  DUP_INX=2
//  
//  DEBUG: TEST2: NEW_WAY=1
//  TEST2: BEFORE UNIQ: 10,20,21,22,30,31,23,24,11
//  TEST2: AFTER UNIQ: 10,20,30,23,11,31,22,24,21
//  DUP_INX=5
//
//  @SEE: look at the "case study" below to understand which the last "AFTER UNIQ" results with that order:
//  TEST2: AFTER UNIQ: 10,20,30,23,11,31,22,24,21
//
//==========================================================================================================
// Case Study: how ptgi::unique() works:
//  Remember we "remove adjacent duplicates".
//  In this example, the input is NOT fully sorted when ptgi:unique() is called.
//
//  I put | separatators, BEFORE UNIQ to illustrate this
//  10  | 20,21,22 |  30,31 |  23,24 | 11
//
//  In example above, 20, 21, 22 are "same" since dividing by 10 gives 2 quotient.
//  And 30,31 are "same", since /10 quotient is 3.
//  And 23, 24 are same, since /10 quotient is 2.
//  And 11 is "group of one" by itself.
//  So there are 5 groups, but the 4th group (23, 24) happens to be equal to group 2 (20, 21, 22)
//  So there are 5 groups, and the 5th group (11) is equal to group 1 (10)
//
//  R = result
//  F = first
//
//  10, 20, 21, 22, 30, 31, 23, 24, 11
//  R    F
//
//  10 is result, and first points to 20, and R != F (10 != 20) so bump R:
//       R
//       F
//
//  Now we hits the "optimized out swap logic".
//  (avoid swap because R == F)
//
//  // now bump F until R != F (integer division by 10)
//  10, 20, 21, 22, 30, 31, 23, 24, 11
//       R   F              // 20 == 21 in 10x
//       R       F              // 20 == 22 in 10x
//       R           F          // 20 != 30, so we do a swap of ++R and F
//  (Now first hits 21, 22, then finally 30, which is different than R, so we swap bump R to 21 and swap with  30)
//  10, 20, 30, 22, 21, 31, 23, 24, 11  // after R & F swap (21 and 30)
//           R       F 
//
//  10, 20, 30, 22, 21, 31, 23, 24, 11
//           R          F           // bump F to 31, but R and F are same (30 vs 31)
//           R               F      // bump F to 23, R != F, so swap ++R with F
//  10, 20, 30, 22, 21, 31, 23, 24, 11
//                  R           F       // bump R to 22
//  10, 20, 30, 23, 21, 31, 22, 24, 11  // after the R & F swap (22 & 23 swap)
//                  R            F      // will swap 22 and 23
//                  R                F      // bump F to 24, but R and F are same in 10x
//                  R                    F  // bump F, R != F, so swap ++R  with F
//                      R                F  // R and F are diff, so swap ++R  with F (21 and 11)
//  10, 20, 30, 23, 11, 31, 22, 24, 21
//                      R                F  // aftter swap of old 21 and 11
//                      R                  F    // F now at last(), so loop terminates
//                          R               F   // bump R by 1 to point to dupPostion (first duplicate in range)
//
//  return R which now points to 31
//==========================================================================================================
// NOTES:
// 1) the #ifdef IMPROVED_STD_UNIQUE_ALGORITHM documents how we have modified the original std::unique().
// 2) I've heavily unit tested this code, including using valgrind(1), and it is *believed* to be 100% defect-free.
//
//==========================================================================================================
// History:
//  130201  dpb dbednar@ptgi.com created
//==========================================================================================================

#ifndef PTGI_UNIQUE_HPP
#define PTGI_UNIQUE_HPP

// Created to solve memory leak problems when calling std::unique() on a vector<Route*>.
// Memory leaks discovered with valgrind and unitTesting.


#include <algorithm>        // std::swap

// instead of std::myUnique, call this instead, where arg3 is a function ptr
//
// like std::unique, it puts the dups at the end, but it uses swapping to preserve original
// vector contents, to avoid memory leaks and duplicate pointers in vector<Object*>.

#ifdef IMPROVED_STD_UNIQUE_ALGORITHM
#error the #ifdef for IMPROVED_STD_UNIQUE_ALGORITHM was defined previously.. Something is wrong.
#endif

#undef IMPROVED_STD_UNIQUE_ALGORITHM
#define IMPROVED_STD_UNIQUE_ALGORITHM

// similar to std::unique, except that this version swaps elements, to avoid
// memory leaks, when vector contains pointers.
//
// Normally the input is sorted.
// Normal std::unique:
// 10 20 20 20 30   30 20 20 10
// a  b  c  d  e    f  g  h  i
//
// 10 20 30 20 10 | 30 20 20 10
// a  b  e  g  i    f  g  h  i
//
// Now GONE: c, d.
// Now DUPS: g, i.
// This causes memory leaks and segmenation faults due to duplicate deletes of same pointer!


namespace ptgi {

// Return the position of the first in range of duplicates moved to end of vector.
//
// uses operator==  of class for comparison
//
// @param [first, last) is a range to find duplicates within.
//
// @return the dupPosition position, such that [dupPosition, end) are contiguous
// duplicate elements.
// IF all items are unique, then it would return last.
//
template <class ForwardIterator>
ForwardIterator unique( ForwardIterator first, ForwardIterator last)
{
    // compare iterators, not values
    if (first == last)
        return last;

    // remember the current item that we are looking at for uniqueness
    ForwardIterator result = first;

    // result is slow ptr where to store next unique item
    // first is  fast ptr which is looking at all elts

    // the first iterator moves over all elements [begin+1, end).
    // while the current item (result) is the same as all elts
    // to the right, (first) keeps going, until you find a different
    // element pointed to by *first.  At that time, we swap them.

    while (++first != last)
    {
        if (!(*result == *first))
        {
#ifdef IMPROVED_STD_UNIQUE_ALGORITHM
            // inc result, then swap *result and *first

//          THIS IS WHAT WE WANT TO DO.
//          BUT THIS COULD SWAP AN ELEMENT WITH ITSELF, UNCECESSARILY!!!
//          std::swap( *first, *(++result));

            // BUT avoid swapping with itself when both iterators are the same
            ++result;
            if (result != first)
                std::swap( *first, *result);
#else
            // original code found in std::unique()
            // copies unique down
            *(++result) = *first;
#endif
        }
    }

    return ++result;
}

template <class ForwardIterator, class BinaryPredicate>
ForwardIterator unique( ForwardIterator first, ForwardIterator last, BinaryPredicate pred)
{
    if (first == last)
        return last;

    // remember the current item that we are looking at for uniqueness
    ForwardIterator result = first;

    while (++first != last)
    {
        if (!pred(*result,*first))
        {
#ifdef IMPROVED_STD_UNIQUE_ALGORITHM
            // inc result, then swap *result and *first

//          THIS COULD SWAP WITH ITSELF UNCECESSARILY
//          std::swap( *first, *(++result));
//
            // BUT avoid swapping with itself when both iterators are the same
            ++result;
            if (result != first)
                std::swap( *first, *result);

#else
            // original code found in std::unique()
            // copies unique down
            // causes memory leaks, and duplicate ptrs
            // and uncessarily moves in place!
            *(++result) = *first;
#endif
        }
    }

    return ++result;
}

// from now on, the #define is no longer needed, so get rid of it
#undef IMPROVED_STD_UNIQUE_ALGORITHM

} // end ptgi:: namespace

#endif

这是我用来测试它的UNIT Test程序:

// QUESTION: in test2, I had trouble getting one line to compile,which was caused  by the declaration of operator()
// in the equal_to Predicate.  I'm not sure how to correctly resolve that issue.
// Look for //OUT lines
//
// Make sure that NOTES in ptgi_unique.hpp are correct, in how we should "cleanup" duplicates
// from both a vector<Integer> (test1()) and vector<Integer*> (test2).
// Run this with valgrind(1).
//
// In test2(), IF we use the call to std::unique(), we get this problem:
//
//  [dbednar@ipeng8 TestSortRoutes]$ ./Main7
//  TEST2: ORIG nums before UNIQUE: 10, 20, 21, 22, 30, 31, 23, 24, 11
//  TEST2: modified nums AFTER UNIQUE: 10, 20, 30, 23, 11, 31, 23, 24, 11
//  INFO: dupInx=5
//  TEST2: uniq = 10
//  TEST2: uniq = 20
//  TEST2: uniq = 30
//  TEST2: uniq = 33427744
//  TEST2: uniq = 33427808
//  Segmentation fault (core dumped)
//
// And if we run valgrind we seen various error about "read errors", "mismatched free", "definitely lost", etc.
//
//  valgrind --leak-check=full ./Main7
//  ==359== Memcheck, a memory error detector
//  ==359== Command: ./Main7
//  ==359== Invalid read of size 4
//  ==359== Invalid free() / delete / delete[]
//  ==359== HEAP SUMMARY:
//  ==359==     in use at exit: 8 bytes in 2 blocks
//  ==359== LEAK SUMMARY:
//  ==359==    definitely lost: 8 bytes in 2 blocks
// But once we replace the call in test2() to use ptgi::unique(), all valgrind() error messages disappear.
//
// 130212   dpb dbednar@ptgi.com created
// =========================================================================================================

#include <iostream> // std::cout, std::cerr
#include <string>
#include <vector>   // std::vector
#include <sstream>  // std::ostringstream
#include <algorithm>    // std::unique()
#include <functional>   // std::equal_to(), std::binary_function()
#include <cassert>  // assert() MACRO

#include "ptgi_unique.hpp"  // ptgi::unique()



// Integer is small "wrapper class" around a primitive int.
// There is no SETTER, so Integer's are IMMUTABLE, just like in JAVA.

class Integer
{
private:
    int num;
public:

    // default CTOR: "Integer zero;"
    // COMPRENSIVE CTOR:  "Integer five(5);"
    Integer( int num = 0 ) :
        num(num)
    {
    }

    // COPY CTOR
    Integer( const Integer& rhs) :
        num(rhs.num)
    {
    }

    // assignment, operator=, needs nothing special... since all data members are primitives

    // GETTER for 'num' data member
    // GETTER' are *always* const
    int getNum() const
    {
        return num;
    }   

    // NO SETTER, because IMMUTABLE (similar to Java's Integer class)

    // @return "num"
    // NB: toString() should *always* be a const method
    //
    // NOTE: it is probably more efficient to call getNum() intead
    // of toString() when printing a number:
    //
    // BETTER to do this:
    //  Integer five(5);
    //  std::cout << five.getNum() << "\n"
    // than this:
    //  std::cout << five.toString() << "\n"

    std::string toString() const
    {
        std::ostringstream oss;
        oss << num;
        return oss.str();
    }
};

// convenience typedef's for iterating over std::vector<Integer>
typedef std::vector<Integer>::iterator      IntegerVectorIterator;
typedef std::vector<Integer>::const_iterator    ConstIntegerVectorIterator;

// convenience typedef's for iterating over std::vector<Integer*>
typedef std::vector<Integer*>::iterator     IntegerStarVectorIterator;
typedef std::vector<Integer*>::const_iterator   ConstIntegerStarVectorIterator;

// functor used for std::unique or ptgi::unique() on a std::vector<Integer>
// Two numbers equal if, when divided by 10 (integer division), the quotients are the same.
// Hence 50..59 are equal, 60..69 are equal, etc.
struct IntegerEqualByTen: public std::equal_to<Integer>
{
    bool operator() (const Integer& arg1, const Integer& arg2) const
    {
        return ((arg1.getNum()/10) == (arg2.getNum()/10));
    }
};

// functor used for std::unique or ptgi::unique on a std::vector<Integer*>
// Two numbers equal if, when divided by 10 (integer division), the quotients are the same.
// Hence 50..59 are equal, 60..69 are equal, etc.
struct IntegerEqualByTenPointer: public std::equal_to<Integer*>
{
    // NB: the Integer*& looks funny to me!
    // TECHNICAL PROBLEM ELSEWHERE so had to remove the & from *&
//OUT   bool operator() (const Integer*& arg1, const Integer*& arg2) const
//
    bool operator() (const Integer* arg1, const Integer* arg2) const
    {
        return ((arg1->getNum()/10) == (arg2->getNum()/10));
    }
};

void test1();
void test2();
void printIntegerStarVector( const std::string& msg, const std::vector<Integer*>& nums );

int main()
{
    test1();
    test2();
    return 0;
}

// test1() uses a vector<Object> (namely vector<Integer>), so there is no problem with memory loss
void test1()
{
    int data[] = { 10, 20, 21, 22, 30, 31, 23, 24, 11};

    // turn C array into C++ vector
    std::vector<Integer> nums(data, data+9);

    // arg3 is a functor
    IntegerVectorIterator dupPosition = ptgi::unique( nums.begin(), nums.end(), IntegerEqualByTen() );

    nums.erase(dupPosition, nums.end());

    nums.erase(nums.begin(), dupPosition);
}

//==================================================================================
// test2() uses a vector<Integer*>, so after ptgi:unique(), we have to be careful in
// how we eliminate the duplicate Integer objects stored in the heap.
//==================================================================================
void test2()
{
    int data[] = { 10, 20, 21, 22, 30, 31, 23, 24, 11};

    // turn C array into C++ vector of Integer* pointers
    std::vector<Integer*> nums;

    // put data[] integers into equivalent Integer* objects in HEAP
    for (int inx = 0; inx < 9; ++inx)
    {
        nums.push_back( new Integer(data[inx]) );
    }

    // print the vector<Integer*> to stdout
    printIntegerStarVector( "TEST2: ORIG nums before UNIQUE", nums );

    // arg3 is a functor
#if 1
    // corrected version which fixes SEGMENTATION FAULT and all memory leaks reported by valgrind(1)
    // I THINK we want to use new C++11 cbegin() and cend(),since the equal_to predicate is passed "Integer *&"

//  DID NOT COMPILE
//OUT   IntegerStarVectorIterator dupPosition = ptgi::unique( const_cast<ConstIntegerStarVectorIterator>(nums.begin()), const_cast<ConstIntegerStarVectorIterator>(nums.end()), IntegerEqualByTenPointer() );

    // DID NOT COMPILE when equal_to predicate declared "Integer*& arg1, Integer*&  arg2"
//OUT   IntegerStarVectorIterator dupPosition = ptgi::unique( const_cast<nums::const_iterator>(nums.begin()), const_cast<nums::const_iterator>(nums.end()), IntegerEqualByTenPointer() );


    // okay when equal_to predicate declared "Integer* arg1, Integer*  arg2"
    IntegerStarVectorIterator dupPosition = ptgi::unique(nums.begin(), nums.end(), IntegerEqualByTenPointer() );
#else
    // BUGGY version that causes SEGMENTATION FAULT and valgrind(1) errors
    IntegerStarVectorIterator dupPosition = std::unique( nums.begin(), nums.end(), IntegerEqualByTenPointer() );
#endif

    printIntegerStarVector( "TEST2: modified nums AFTER UNIQUE", nums );
    int dupInx = dupPosition - nums.begin();
    std::cout << "INFO: dupInx=" << dupInx <<"\n";

    // delete the dup Integer* objects in the [dupPosition, end] range
    for (IntegerStarVectorIterator iter = dupPosition; iter != nums.end(); ++iter)
    {
        delete (*iter);
    }

    // shrink the vector
    // NB: the Integer* ptrs are NOT followed by vector::erase()
    nums.erase(dupPosition, nums.end());


    // print the uniques, by following the iter to the Integer* pointer
    for (IntegerStarVectorIterator iter = nums.begin(); iter != nums.end();  ++iter)
    {
        std::cout << "TEST2: uniq = " << (*iter)->getNum() << "\n";
    }

    // remove the unique objects from heap
    for (IntegerStarVectorIterator iter = nums.begin(); iter != nums.end();  ++iter)
    {
        delete (*iter);
    }

    // shrink the vector
    nums.erase(nums.begin(), nums.end());

    // the vector should now be completely empty
    assert( nums.size() == 0);
}

//@ print to stdout the string: "info_msg: num1, num2, .... numN\n"
void printIntegerStarVector( const std::string& msg, const std::vector<Integer*>& nums )
{
    std::cout << msg << ": ";
    int inx = 0;
    ConstIntegerStarVectorIterator  iter;

    // use const iterator and const range!
    // NB: cbegin() and cend() not supported until LATER (c++11)
    for (iter = nums.begin(), inx = 0; iter != nums.end(); ++iter, ++inx)
    {
        // output a comma seperator *AFTER* first
        if (inx > 0)
            std::cout << ", ";

        // call Integer::toString()
        std::cout << (*iter)->getNum();     // send int to stdout
//      std::cout << (*iter)->toString();   // also works, but is probably slower

    }

    // in conclusion, add newline
    std::cout << "\n";
}

我不明白这里的理由。因此,如果您有一个指针容器,并且想要删除重复项,这将如何影响指针所指向的对象?不会发生内存泄漏,因为至少有一个指针(指向该容器中的一个)指向它们。好吧,好吧,我想您的方法可能需要某些奇怪的重载运算符或奇怪的比较函数,这些优点确实需要特别考虑。
kccqzy13年

不知道我是否理解你的意思。以vector <int *>的简单情况为例,其中4个指针指向整数{1、2、2、3}。它已排序,但是在您调用std :: unique之后,这4个指针是指向整数{1、2、3、3}的指针。现在,您有两个完全相同的指向3的指针,因此,如果您调用delete,它将执行重复删除。坏!其次,请注意第二个2丢失了,这是内存泄漏。

kccqzy,以下示例程序可帮助您更好地理解我的答案:
joe

@joe:即使std::unique您拥有[ 1、2、3、2 ],也无法在2上调用delete,因为那样会留下指向2的悬挂指针!=>根本不要在newEnd = std::unique和之间的元素上调用delete ,std::end因为您仍然在其中有指向这些元素的指针[std::begin, newEnd)
MFH,2014年

2
@ArneVogel:也许对于“正常工作”的琐碎值。调用uniquea 毫无意义vector<unique_ptr<T>>,因为矢量可以包含的唯一重复值是nullptr
Ben Voigt

2

使用Ranges库(在C ++ 20中提供),您可以简单地使用

action::unique(vec);

请注意,它实际上删除了重复的元素,而不仅仅是移动它们。


1

关于alexK7基准测试。我尝试了这些方法,并得到了相似的结果,但是当值的范围为100万时,使用std :: sort(f1)和使用std :: unordered_set(f5)的情况产生的时间相似。当值的范围是1000万时,f1快于f5。

如果值的范围是有限的并且值是unsigned int,则可以使用std :: vector,其大小对应于给定的范围。这是代码:

void DeleteDuplicates_vector_bool(std::vector<unsigned>& v, unsigned range_size)
{
    std::vector<bool> v1(range_size);
    for (auto& x: v)
    {
       v1[x] = true;    
    }
    v.clear();

    unsigned count = 0;
    for (auto& x: v1)
    {
        if (x)
        {
            v.push_back(count);
        }
        ++count;
    }
}

1

sort(v.begin(),v.end()),v.erase(unique(v.begin(),v,end()),v.end());


1

如果您正在寻找性能并使用std::vector,我建议使用此文档链接提供的一种。

std::vector<int> myvector{10,20,20,20,30,30,20,20,10};             // 10 20 20 20 30 30 20 20 10
std::sort(myvector.begin(), myvector.end() );
const auto& it = std::unique (myvector.begin(), myvector.end());   // 10 20 30 ?  ?  ?  ?  ?  ?
                                                                   //          ^
myvector.resize( std::distance(myvector.begin(),it) ); // 10 20 30

cplusplus.com绝不是官方文档。
伊利亚·波波夫

0
std::set<int> s;
std::for_each(v.cbegin(), v.cend(), [&s](int val){s.insert(val);});
v.clear();
std::copy(s.cbegin(), s.cend(), v.cbegin());

1
也许在清除向量后重新调整向量的大小,以便在构建向量时仅分配1个内存。也许更喜欢使用std :: move而不是std :: copy将int移到向量中,而不是复制它们,因为以后将不需要该集合。
YoungJohn

0

如果你希望修改向量(删除,排序),那么你可以用牛顿库,在算法子库有一个函数调用,copy_single

template <class INPUT_ITERATOR, typename T>
    void copy_single( INPUT_ITERATOR first, INPUT_ITERATOR last, std::vector<T> &v )

这样你就可以:

std::vector<TYPE> copy; // empty vector
newton::copy_single(first, last, copy);

其中copy是要回唯一元素的副本的向量。但请记住,您push_back了元素,并且您没有创建新的向量

无论如何,这是更快的方法,因为您无需擦除()元素(这需要花费大量时间,但由于重新分配而需要pop_back()时除外)

我做了一些实验,速度更快。

另外,您可以使用:

std::vector<TYPE> copy; // empty vector
newton::copy_single(first, last, copy);
original = copy;

有时还是更快。


1
此函数在标准库中以形式存在unique_copy
LF

0

来自以下网址的更易理解的代码:https//en.cppreference.com/w/cpp/algorithm/unique

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cctype>

int main() 
{
    // remove duplicate elements
    std::vector<int> v{1,2,3,1,2,3,3,4,5,4,5,6,7};
    std::sort(v.begin(), v.end()); // 1 1 2 2 3 3 3 4 4 5 5 6 7 
    auto last = std::unique(v.begin(), v.end());
    // v now holds {1 2 3 4 5 6 7 x x x x x x}, where 'x' is indeterminate
    v.erase(last, v.end()); 
    for (int i : v)
      std::cout << i << " ";
    std::cout << "\n";
}

输出:

1 2 3 4 5 6 7

0
void removeDuplicates(std::vector<int>& arr) {
    for (int i = 0; i < arr.size(); i++)
    {
        for (int j = i + 1; j < arr.size(); j++)
        {
            if (arr[i] > arr[j])
            {
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
    std::vector<int> y;
    int x = arr[0];
    int i = 0;
    while (i < arr.size())
    {
        if (x != arr[i])
        {
            y.push_back(x);
            x = arr[i];
        }
        i++;
        if (i == arr.size())
            y.push_back(arr[i - 1]);
    }
    arr = y;
}

2
欢迎来到StackOverflow!请编辑您的问题,以解释您的代码如何工作,以及为什么它与其他答案同等或更好。这个问题已有十多年的历史了,并且已经有了很多很好的解释。如果没有您的解释,它就没有用,并且很有可能被拒绝或删除。
Das_Geek,

-1

这是std :: unique()发生重复删除问题的示例。在LINUX机器上,程序崩溃。阅读评论以获取详细信息。

// Main10.cpp
//
// Illustration of duplicate delete and memory leak in a vector<int*> after calling std::unique.
// On a LINUX machine, it crashes the progam because of the duplicate delete.
//
// INPUT : {1, 2, 2, 3}
// OUTPUT: {1, 2, 3, 3}
//
// The two 3's are actually pointers to the same 3 integer in the HEAP, which is BAD
// because if you delete both int* pointers, you are deleting the same memory
// location twice.
//
//
// Never mind the fact that we ignore the "dupPosition" returned by std::unique(),
// but in any sensible program that "cleans up after istelf" you want to call deletex
// on all int* poitners to avoid memory leaks.
//
//
// NOW IF you replace std::unique() with ptgi::unique(), all of the the problems disappear.
// Why? Because ptgi:unique merely reshuffles the data:
// OUTPUT: {1, 2, 3, 2}
// The ptgi:unique has swapped the last two elements, so all of the original elements in
// the INPUT are STILL in the OUTPUT.
//
// 130215   dbednar@ptgi.com
//============================================================================

#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>

#include "ptgi_unique.hpp"

// functor used by std::unique to remove adjacent elts from vector<int*>
struct EqualToVectorOfIntegerStar: public std::equal_to<int *>
{
    bool operator() (const int* arg1, const int* arg2) const
    {
        return (*arg1 == *arg2);
    }
};

void printVector( const std::string& msg, const std::vector<int*>& vnums);

int main()
{
    int inums [] = { 1, 2, 2, 3 };
    std::vector<int*> vnums;

    // convert C array into vector of pointers to integers
    for (size_t inx = 0; inx < 4; ++ inx)
        vnums.push_back( new int(inums[inx]) );

    printVector("BEFORE UNIQ", vnums);

    // INPUT : 1, 2A, 2B, 3
    std::unique( vnums.begin(), vnums.end(), EqualToVectorOfIntegerStar() );
    // OUTPUT: 1, 2A, 3, 3 }
    printVector("AFTER  UNIQ", vnums);

    // now we delete 3 twice, and we have a memory leak because 2B is not deleted.
    for (size_t inx = 0; inx < vnums.size(); ++inx)
    {
        delete(vnums[inx]);
    }
}

// print a line of the form "msg: 1,2,3,..,5,6,7\n", where 1..7 are the numbers in vnums vector
// PS: you may pass "hello world" (const char *) because of implicit (automatic) conversion
// from "const char *" to std::string conversion.

void printVector( const std::string& msg, const std::vector<int*>& vnums)
{
    std::cout << msg << ": ";

    for (size_t inx = 0; inx < vnums.size(); ++inx)
    {
        // insert comma separator before current elt, but ONLY after first elt
        if (inx > 0)
            std::cout << ",";
        std::cout << *vnums[inx];

    }
    std::cout << "\n";
}

PS:我还运行了“ valgrind ./Main10”,而valgrind发现没有问题。我强烈建议所有使用LINUX的C ++程序员使用此高效工具,尤其是如果您要编写必须运行24x7且不会泄漏或崩溃的实时应用程序时!

std :: unique问题的核心可以由以下语句总结:“ std :: unique返回未指定状态的重复项”!为什么标准委员会会这样做,我永远不会知道。承诺成员..任何意见???

1
是的,“ std :: unique返回未指定状态的重复项”。因此,根本不要依靠已经“唯一”的数组来手动管理内存!最简单的方法是使用std :: unique_ptr而不是原始指针。
alexk7 2014年

这似乎是对另一个答案的回应;它不能回答问题(其中vector包含整数,不包含指针,并且未指定比较器)。
Toby Speight

-2
void EraseVectorRepeats(vector <int> & v){ 
TOP:for(int y=0; y<v.size();++y){
        for(int z=0; z<v.size();++z){
            if(y==z){ //This if statement makes sure the number that it is on is not erased-just skipped-in order to keep only one copy of a repeated number
                continue;}
            if(v[y]==v[z]){
                v.erase(v.begin()+z); //whenever a number is erased the function goes back to start of the first loop because the size of the vector changes
            goto TOP;}}}}

这是我创建的一个函数,可用于删除重复项。所需的头文件是just <iostream><vector>

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