在调用unique之前需要对它进行排序，因为unique只删除相邻的重复项。

编辑:38秒……

效率是一个复杂的概念。有时间和空间的考虑，以及一般的测量(你只能得到模糊的答案，如O(n))和特定的(例如冒泡排序可以比快速排序快得多，这取决于输入特征)。

如果你有相对较少的副本，那么排序，然后唯一和擦除似乎是要走的路。如果您有相对较多的副本，则从向量创建一个集合，并让它完成繁重的工作，可以轻松击败它。

也不要只关注时间效率。Sort+unique+erase操作在O(1)空间，而set构造操作在O(n)空间。而且这两种方法都不能直接用于map-reduce并行化(对于真正庞大的数据集)。

你可以这样做:

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

大部分答案似乎是使用O(nlogn)，但使用unordered_set，我们可以将其减少到O(n)。我看到了一些使用集合的解决方案，但我发现了这个解决方案，使用集合和迭代器似乎更优雅。

using Intvec = std::vector<int>;

void remove(Intvec &v) {
    // creating iterator starting with beginning of the vector 
    Intvec::iterator itr = v.begin();
    std::unordered_set<int> s;
    // loops from the beginning to the end of the list 
    for (auto curr = v.begin(); curr != v.end(); ++curr) {
        if (s.insert(*curr).second) { // if the 0 curr already exist in the set
            *itr++ = *curr; // adding a position to the iterator 
        }
    }
    // erasing repeating positions in the set 
    v.erase(itr, v.end());
}

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());

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";
}