Sum of three numbers

This problem uses a double pointer Algorithm , How to use double pointers when there are three numbers , We can fix a number first i, Then double pointer j and k.

The key point of weight removal is designed in this problem ：

1. Guarantee i<k<j,, Why? , Here is an example to know , for instance ,i take -1,j take 0,k take 1. If we can't meet the conditions, we can take i take -1,j take 1,k take 0. Does this repeat .

2. If i If the numbers taken twice are the same, there will be repetition

For example, the array is 1 1 1 1 1 1 -2;

here i Take the first one 1, and i Take the second 1 It's all repetitive , So we need to be i Skip directly when traversing to the same number .

Ideas ：

We use double pointers in this problem , Double pointers need to be ordered , We can arrange the order in advance , then i and j All have to be de weighted , Next is the double pointer operation .

step ：

1. Create an answer array

2. Loop defines the first pointer , Then repeat the operation every time ,

3. Carry out the second cycle ,j=i+1,k=nums.size()-1,j<k;j++; And carry out the weight removal operation again

4. If you can, you can ,k--, Finally, judge whether the addition of three numbers is 0, If 0 Just save it in the answer to get .

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
       vector<vector<int>>res;
       sort(nums.begin(),nums.end());
       for(int i=0;i<nums.size();i++){
           if(i&&nums[i-1]==nums[i])continue;
           for(int j=i+1,k=nums.size()-1;j<k;j++){
               if(j>i+1&&nums[j]==nums[j-1])continue;
               while(j<k-1&&nums[i]+nums[j]+nums[k-1]>=0)k--;
               if(nums[i]+nums[j]+nums[k]==0)
                   res.push_back({nums[i],nums[j],nums[k]});
           }
       }
       return res;

    }
};