15. 3Sum

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.

Note:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:(-1, 0, 1)(-1, -1, 2)

O(n*n)
排好序之后
三个指针,i,j,k
遍历i,先固定i
然后j = i+1,k = length - 1;
j,k两个指针逼近，看加起来是否a[j] + a[k] = - a[i]
如果大于,k–
如果小于,j++
如果等于则把j,k分别移到不同的位置
再把i移到和之前不同的地方，重新开始

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;}