LeetCode 3Sum

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)

class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int> > r;        int i, j, n = num.size();        bool flag = true;                sort(num.begin(),num.end());        for(int k = 0; k <= n - 3; k++) {            if(num[k] == num[k+1] && num[k+1] == num[k+2]) {                if(num[k] == 0 && flag) {                    vector<int> a(3,0);                    r.push_back(a);                    flag = false;                }                else continue;            }            else if(num[k] == num[k+1]) {                for(i = k + 2; i <= n - 1; i++) {                    if(num[i]+num[k]+num[k+1]==0) {                        vector<int> a(3,0);                        a[0] = num[k];                        a[1] = num[k+1];                        a[2] = num[i];                        r.push_back(a);                        break;                    }                }            }            else {                i = k + 1;                j = n - 1;                while(i < j) {                    if(num[i]+num[j]+num[k]==0) {                        vector<int> a(3,0);                        a[0] = num[k];                        a[1] = num[i];                        a[2] = num[j];                        r.push_back(a);                        while(num[i] == num[i+1]) {                                i++;                        }                        i++;                    }else if(num[i]+num[j]+num[k]<0) i++;                    else j--;                }            }        }        return r;    }};

思路：与 3Sum Closest 思路相似。但为避免出现重复答案，做了大量工作。