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)

. 1. 首选排序N*log(N);
  2. 第一重循环从第一个元素开始，每次想后递增到一个不重复的下一个节点（因为内部循环已经把最小的值为当前值得所有可能都遍历过了）
  3. 内部循环从两头向内遍历，根据3个元素的加和来决定想后移动前偏移还是向前移动和后偏移，同时要求当满足和为0的时候，至少有一个值和前一个满足题意的值不同，这样就去掉了所有重复的可能


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class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int>> ret;        int cnt = num.size();        if (cnt < 3) return ret;                sort(num.begin(), num.end());        int preval = 0;        for (int i = 0; i < cnt - 2; i++) {            if (i > 0 && num[i] == preval) continue;            preval = num[i];            int j = i + 1, k = cnt - 1;            int cj = INT_MIN, ck = INT_MIN;            while (j < k) {                int cval = preval + num[j] + num[k];                if (cval == 0) {                     if (cj != num[j] || ck != num[k]) {                         ret.push_back(vector<int>(3, preval));                         cj = num[j];                         ck = num[k];                         ret.back()[1] = cj;                         ret.back()[2] = ck;                     }                                          j++;                     k--;                } else if (cval > 0) {                    k--;                } else {                    j++;                }            }        }                return ret;    }};