Leetcode (question 15) - sum of three numbers

One 、 Title Description

 Give you one containing  n  Array of integers  nums, Judge  nums  Are there three elements in  a,b,c , bring  a + b + c = 0 ？ Please find out all and for  0  Triple without repetition .
 Be careful ： Answer cannot contain duplicate triples .

Two 、 Example

 Example 1 
 Input ：nums = [-1,0,1,2,-1,-4]
 Output ：[[-1,-1,2],[-1,0,1]]

 Example 2 
 Input ：nums = []
 Output ：[]

 Example 3 
 Input ：nums = [0]
 Output ：[]

3、 ... and 、 Ideas
1、 First, let's determine the conditions that are not met , If the length of the array is less than three , It returns an empty array directly .
2、 If the length of the array is greater than or equal to three , Then sort the array , In order from small to large .
3、 At this point, we judge the first element of the array , If the size of the first element is greater than 0, It returns an empty array directly .
4、 We determine the position of three numbers , Suppose the three numbers are a,b,c, also a <= b <= c.
5、 At this point, we will traverse the elements of the array , From the first , Traverse to the penultimate , At this point, the element traversed is assumed to be a Elements .
6、 In the case of each element traversed above , hypothesis b Elements by a The next element of the element ,c Elements For the last element of the array .
7、 And then determine a + b + c Is it 0, If 0, Then put it into the array , If it is greater than 0, Will c Elements Move one bit to the left , If it is less than 0, Will b Elements Move one bit to the right .
8、 There is a condition in the title , Answer cannot contain duplicate triples . So we should de duplicate it . There are three situations (1): If a Element is equal to its previous element , Should be directly continue,(2): If b Element is equal to its next element , Then its subscript should be shifted one bit to the right .(3): If c Element is equal to its previous element , The subscript should be shifted one bit to the left .
Four 、 Code

/** * @param {number[]} nums * @return {number[][]} */
var threeSum = function(nums) { 
    
    let result = []
    if(nums.length < nums) return result
    let length = nums.length
    nums.sort((a, b) => a -b)
    for(let i = 0; i < length - 2; i++) { 
    
        let cur = nums[i]
        if(cur > 0) return result
        if(i > 0 && cur === nums[i - 1]) continue
        let left = i + 1
        let right = length - 1
        while(left < right) { 
    
            let sum = cur + nums[left] + nums[right]
            if(sum === 0) { 
    
                result.push([cur, nums[left], nums[right]])
                while(nums[left] === nums[left + 1]) left++
                while(nums[right] === nums[right] - 1) right--
                left++
                right--
            }else if(sum > 0) { 
    
                right--
            }else { 
    
                left++
            }
        }          
    }
    return result
};

5、 ... and 、 summary