[Leetcode] 3sum 三数和

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)

题意：给定值0，使三数之和等于0，返回所有情况。

思路：这题的思路和3sum closest类似。关键在于去重。代码如下：

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) 
    {
        vector<vector<int>> res;
        sort(num.begin(),num.end());
        if(num.size()<3)    return res;

        for(int i=0;i<num.size()-2;++i)
        {
            if(num[i]>0)   break;  //最小值大于0，返回空
            if(i>0&&num[i]==num[i-1]) continue;   //去重
            int l=i+1,r=num.size()-1;
            while(l<r)
            {
                int sum=num[i]+num[l]+num[r];
                if(sum==0)
                {
                    res.push_back({num[i], num[l], num[r]});
                    //去重
                    while(++l<r&&num[l-1]==num[l])
                        ;
                    while(--r>l&&num[r]==num[r+1])
                       ;
                }
                else if(sum<0)
                    ++l;
                else
                    r--;

            }
        }
        return res;    
    }
};