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;
}

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());

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。

我同意R. Pate和Todd Gardner的观点;std::set在这里可能是个好主意。即使你在使用向量时遇到了困难，如果你有足够多的副本，你最好创建一个集合来做这些肮脏的工作。

让我们来比较三种方法:

用向量，sort + unique

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

转换为set(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set(使用构造函数)

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

下面是它们在重复数量变化时的表现:

总结:当副本的数量足够大时，实际上更快地将数据转换为一个集合，然后将数据转储回一个向量。

出于某种原因，手动进行set转换似乎比使用set构造函数更快——至少在我使用的随机数据上是这样。

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";
}

你可以这样做:

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