排序+对撞指针（15. 3Sum ）

以leetcode题目为例：

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

For example, given array S = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]]

#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;vector<vector<int> > threeSum(vector<int> &num) {    vector<vector<int> > res;    std::sort(num.begin(), num.end()); // 双指针对撞    for (int i = 0; i < num.size(); i++) {        int target = -num[i];        // 在[i+1..n-1]中用双指针对撞法，找两个数的值为target        // 因为sort后，num[i]最小，因此取负号最大，由此可以确定在上述区间查找        int front = i + 1;        int back = num.size() - 1;        // 对撞指针终止于指针对撞        while (front < back) {            int sum = num[front] + num[back];            // sum小的话就 front往后移            if (sum < target)                front++;            // sum小的话就 back往前移            else if (sum > target)                back--;            // 找到一组符合条件，存下来            else {                vector<int> triplet(3, 0);                triplet[0] = num[i];                triplet[1] = num[front];                triplet[2] = num[back];                res.push_back(triplet);                // 重复的数字只取一个                // Processing duplicates of Number 2                while (front < back && num[front] == triplet[1]) front++;                // Processing duplicates of Number 3                while (front < back && num[back] == triplet[2]) back--;            }        }        // Processing duplicates of Number 1        while (i + 1 < num.size() && num[i + 1] == num[i])            i++;    }    return res;}