15.3Sum

思路 

2sum的升级版

先做一个排序

从第一个开始遍历

从后面的数字找到两个数的和等于第一个的负数

因为排过序之后

取两边方向加到中间即可

关键在于去除冗余

所以要加上一些个判断

vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int> > res;    std::sort(num.begin(), num.end());    for (int i = 0; i < num.size(); i++) {                int target = -num[i];        int front = i + 1;        int back = num.size() - 1;        while (front < back) {            int sum = num[front] + num[back];                        // Finding answer which start from number num[i]            if (sum < target)                front++;            else if (sum > target)                back--;            else {                vector<int> triplet(3, 0);                triplet[0] = num[i];                triplet[1] = num[front];                triplet[2] = num[back];                res.push_back(triplet);                                // Processing duplicates of Number 2                // Rolling the front pointer to the next different number forwards                while (front < back && num[front] == triplet[1]) front++;                // Processing duplicates of Number 3                // Rolling the back pointer to the next different number backwards                while (front < back && num[back] == triplet[2]) rear--;            }                    }        // Processing duplicates of Number 1        while (i + 1 < num.size() && num[i + 1] == num[i])             i++;    }        return res;    }