【leetcode】3Sum (medium)

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.

 

思路：

把最小值从头到尾循环，中间值和最大值两边收缩。

我写的时候是在最后去重复，总是超时。后来看了人家的答案，发现可以每次对最小、中间、最大值去重。就可以AC了

#include <iostream>

#include <vector>

#include <algorithm>

#include <queue>

#include <stack>

using namespace std;



class Solution {

public:

    vector<vector<int> > threeSum(vector<int> &num) {

        vector<vector<int> > ans;

        if(num.size() < 3)

        {

            return ans;

        }

        int small = 0;

        int big = num.size() - 1;

        int mid = 0;

        int sum = 0;

        sort(num.begin(), num.end());

        for(small = 0; small < num.size() - 2; /*small++*/) //最小数字循环  中间与最大数字两边收缩

        {

            mid = small + 1;

            big = num.size() - 1;

            while(mid < big)

            {

                sum = num[small] + num[mid] + num[big];

                if(sum > 0)

                {

                    big--;

                }

                else if(sum < 0)

                {

                    mid++;

                }

                else

                {

                    vector<int> v;

                    v.push_back(num[small]);

                    v.push_back(num[mid]);

                    v.push_back(num[big]);

                    ans.push_back(v);

                    do { mid++; }while (mid < big && num[mid - 1] == num[mid]); //注意！！

                    do { big--; }while (mid < big && num[big + 1] == num[big]);//注意！！

                }

            }

            do{ small++; }while (small < num.size() - 1 && num[small - 1] == num[small]);//注意！！

        }

        return ans;

    }

};



int main()

{

    Solution s;

    vector<int> num;

    num.push_back(-4);

    num.push_back(-1);

    num.push_back(-1);

    num.push_back(-1);

    num.push_back(-1);

    num.push_back(0);

    num.push_back(0);

    num.push_back(0);

    num.push_back(1);

    num.push_back(2);



    vector<vector<int>> ans = s.threeSum(num);



    return 0;





}