LeetCode 15. 3Sum three and the number of

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The key solution is to remove repeated, map marking out.

Japanese sentence, if the length is less than the array $3$ , return $[\ ]$。

Sort the array.

The first digit enumeration $nums[i]$ , the double pointer to find a two-digit $nums[L]$ and $nums[R]$

• If the $nums[i]>0$ : as it has been sorted, so there can not be equal to three and the number of $0$ , exit.
• For repeating elements: skipped, to avoid duplication Solutions

Make a left pointer $L=i+1$ , the right pointer $R=n−1$ , when $L<R$ , the execution cycle:

• when $nums[i]+nums[L]+nums[R]==0$ , execution cycle, and determines whether or not the left boundary and the right boundary position of the next repeat, deduplication solution. And at the same time $L,R$ to the next position, looking for new solutions
• If greater than $0$ , indicating $nums[R]$ is too large, $R$ Left
• If less than $0$ , indicating $nums[L]$ is too small, $L$ right

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        
        for(int i = 0; i < n; ++i)
        {
            if(nums[i] > 0)	break;	//若此处大于0，后面不可能找到负数了
            if(i > 0 && nums[i] == nums[i-1])   continue;	//去重 
            
            int pivot = -nums[i];
            int L = i + 1, R = n - 1;
            
            while(L<R)
            {
                if(nums[L]+nums[R]==pivot)
                {
                    ans.push_back({nums[i],nums[L],nums[R]});
                    ++L;
					--R; 
                    while(L < R && nums[L] == nums[L-1])
                    	++L;
                    while(L < R && nums[R] == nums[R+1])
                    	--R;
                }
                else if(nums[L]+nums[R] < pivot)
                {
                    ++L;
                }
                else
                {
                    --R;
                }
            }
        }
        return ans;
    }
};