Leetcode 3Sum, leetcode3sum

Leetcode 3Sum, leetcode3sum

We assume that the three numbers of indexes are a, B, and c.

Duplicate production may occur in the following situations:

1. The first number leads to repetition, num [a] = num [A-1], followed by the same combination of B, c will lead to repeated answers

2. Repeat results from the second and third numbers only after an answer is found. Therefore, you only need to confirm that B ++, c -- is a different number.

    vector<vector<int>> threeSum(vector<int> &num) {        int b, c, n = num.size();        vector<vector<int>> result;        sort(num.begin(), num.end());        for(int i = 0; i < n-2; i++){            if(i > 0 && num[i] == num[i-1])                continue;            b = i+1;            c = n-1;            while(b < c){                if(num[i] + num[b] + num[c] < 0)                    b++;                else if(num[i] + num[b] + num[c] > 0)                    c--;                else {                    vector<int> tmp;                    tmp.push_back(num[i]);                    tmp.push_back(num[b]);                    tmp.push_back(num[c]);                    result.push_back(tmp);                    b++;c--;                    while(num[b] == num[b-1])           // avoid duplicates                        b++;                    while(num[c] == num[c+1])                        c--;                }            }        }        return result;    }