[algorithm] sword finger offer2 golang interview question 7: 3 numbers with 0 in the array

[ Algorithm ] The finger of the sword offer2 golang Interview questions 7： Array and 0 Of 3 A digital

subject 1:

Given an inclusion n Array of integers nums, Judge nums Are there three elements in a ,b ,c , bring a + b + c = 0 ？ Please find all and for 0 And No repetition Triple of .

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 ：[]

Tips ：

0 <= nums.length <= 3000
-105 <= nums[i] <= 105

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/1fGaJU
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas 1:

// Ideas :  Double pointer 
// First, sort the array 
// Fix a number nums[i], other 2 The sum of the numbers is -nums[i] that will do , It's a double pointer problem 
// duplicate removal ,i,j If you encounter duplicate numbers, you can skip them and go to the duplicate 

Code

func threeSum(nums []int) [][]int {
    
    // Ideas :  Double pointer 
    // First, sort the array 
    // Fix a number nums[i], other 2 The sum of the numbers is -nums[i] that will do , It's a double pointer problem 
    // duplicate removal ,i,j If you encounter duplicate numbers, you can skip them and go to the duplicate 

    res := make([][]int,0)
    // Processing parameters 
    if len(nums) < 3 {
    
        return res
    }

    // Sort 
    sort.Slice(nums,func(i,j int) bool{
    
        return nums[i] < nums[j]
    })

    // Fix a number  nums[i]
    for i := 0; i < len(nums) -2 ; {
    
        // Double pointer 
        j,k := i+1,len(nums) -1 
        for j < k {
    
            temp := nums[i] + nums[j] + nums[k]
            if temp == 0 {
    
                res = append(res,[]int{
    nums[i],nums[j],nums[k]})
                // duplicate removal 
                jVal := nums[j]
                for j < k && jVal == nums[j]{
    
                    j ++
                }
            }else if temp < 0 {
    
                j ++
            }else {
    
                k --
            }
        }
        // duplicate removal 
        iVal := nums[i]
        for  i < len(nums) && iVal == nums[i]{
    
            i ++
        }
    }
    return res
}

test