[LeetCode]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)

 思考：仿照在2Sum，增加一个循环遍历数组，复杂度O(n²)。一直纠结于是否存在O(nlogn)的解法？最初思路是在2Sum的基础上，二分搜索(0-num[i]-num[j])。未能解决搜索成功后i，j的变化。

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > res;
        vector<int> ans;
        sort(num.begin(),num.end());
        int len=num.size();
        for(int i=0;i<len-2;i++)
        {
            if(i&&num[i]==num[i-1]) continue;
            int left=i+1;
            int right=len-1;
            while(left<right)
            {
                if(left>i+1&&num[left]==num[left-1]) 
                {
                    left++;
                    continue;
                }
                if(right<len-1&&num[right]==num[right+1]) 
                {
                    right--;
                    continue;
                }
                if(num[i]+num[left]+num[right]==0)
                {
                    ans.clear();
                    ans.push_back(num[i]);
                    ans.push_back(num[left]);
                    ans.push_back(num[right]);
                    res.push_back(ans);
                    left++;
                    right--;
                }
                else if(num[i]+num[left]+num[right]>0) right--;
                else left++;
            }
        }
        return res;
    }
};