Unique只删除连续的重复元素(这是它在线性时间内运行所必需的)，因此应该首先执行排序。调用unique后，它将保持排序。

如果你的类很容易转换为int型，并且你有一些内存， Unique可以在没有排序的情况下完成，而且速度快得多:

#include <vector>
#include <stdlib.h>
#include <algorithm>
int main (int argc, char* argv []) {
  //vector init
  std::vector<int> v (1000000, 0);
  std::for_each (v.begin (), v.end (), [] (int& s) {s = rand () %1000;});
  std::vector<int> v1 (v);
  int beg (0), end (0), duration (0);
  beg = clock ();
  {
    std::sort (v.begin (), v.end ());
    auto i (v.begin ());
    i = std::unique (v.begin (), v.end ());
    if (i != v.end ()) v.erase (i, v.end ());
  }
  end = clock ();
  duration = (int) (end - beg);
  std::cout << "\tduration sort + unique == " << duration << std::endl;

  int n (0);
  duration = 0;
  beg = clock ();
  std::for_each (v1.begin (), v1.end (), [&n] (const int& s) {if (s >= n) n = s+1;});
  std::vector<int> tab (n, 0);
  {
    auto i (v1.begin ());
    std::for_each (v1.begin (), v1.end (), [&i, &tab] (const int& s) {
      if (!tab [s]) {
        *i++ = s;
        ++tab [s];
      }
    });
    std::sort (v1.begin (), i);
    v1.erase (i, v1.end ());
  }
  end = clock ();
  duration = (int) (end - beg);
  std::cout << "\tduration unique + sort == " << duration << std::endl;
  if (v == v1) {
    std::cout << "and results are same" << std::endl;
  }
  else {
    std::cout << "but result differs" << std::endl;
  }  
}

典型结果: Duration sort + unique == 38985 持续时间唯一+排序== 2500 结果是一样的

下面是使用std::unique()出现重复删除问题的示例。在LINUX机器上，程序崩溃。详情请阅读评论。

// 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";
}

unique只适用于重复元素的连续运行，所以你最好先排序。但是，它是稳定的，所以你的向量是有序的。

你可以这样做:

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

关于alexK7基准测试。我尝试了它们，得到了类似的结果，但是当值的范围为100万时，使用std::sort (f1)和使用std::unordered_set (f5)的情况产生类似的时间。当取值范围为1000万时，f1比f5快。

如果值的范围是有限的，并且值是无符号int，则可以使用std::vector，其大小对应于给定的范围。代码如下:

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;
    }
}