[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;

    }

};