Leetcode 15. Sum（二分或者暴力或者哈希都可以）

15. 3Sum

Medium

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

题解： 三个数的和，思路就是求任意两个数的和x，然后O(n^2),再用O(1)的搜索搜索-x是否再数组内就可以了，O(1)搜索用哈希就行，或者用O(longn）的二分的方法也可以接收，我用的自带的find函数，最后也过了

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        multiset<int> left;
        multiset<int> right;
        set<vector<int>> ans;
        int fl = 0;
        for(int i = 0;i < nums.size(); i++){
            if(nums[i] == 0) fl++;
            else if(nums[i]<0) left.insert(nums[i]);
            else if(nums[i]>0) right.insert(nums[i]);
        }
        set<int>:: iterator lit;
        set<int>:: iterator rit;
        for(lit = left.begin(); lit!=left.end(); lit++){
            rit = lit;
            rit++;
            for(; rit != left.end(); rit++){
                int tt = (*lit)+(*rit);
                if(right.find(-tt)!=right.end()){
                    vector<int> New;
                    New.push_back( (*lit));New.push_back( (*rit));New.push_back( -tt);
                    ans.insert(New);
                }
            }
        }
        for(rit = right.begin(); rit!=right.end(); rit++){
            lit = rit;
            lit++;
            for(; lit != right.end(); lit++){
                int tt = (*lit)+(*rit);
                if(left.find(-tt)!=left.end()){
                    vector<int> New;
                    New.push_back( -tt);New.push_back( (*rit));New.push_back( (*lit));
                    ans.insert(New);
                }
            }
        }
        if(fl>=1){
            set<int>:: iterator it;
            for(it = left.begin(); it!=left.end(); it++){
                if(right.find(-(*it))!=right.end()){
                    vector<int> New;
                    New.push_back((*it));New.push_back(0);New.push_back(-(*it));
                    ans.insert(New);
                }
            }
        }
        if(fl>=3){
            vector<int> New;
            New.push_back(0);New.push_back(0);New.push_back(0);
            ans.insert(New);
        }
        vector<vector<int>> finalans;
        set<vector<int>>:: iterator i;
        for(i = ans.begin(); i!=ans.end(); i++){
            vector<int> t = (*i);
            finalans.push_back(t);
        }
        return finalans;
    }
};