Force deduction brush questions, sum of three numbers

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 .

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 ：[]

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/3sum
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas ：

（1） First, the array is null Or less than 3 Array of , Directly back to the set .

（2） Sort the array ,

（3） Traversing sorted arrays , Define left and right pointers , Left pointer L by i+1, Right pointer R Is array length l-1.

（4） Ergodic condition ,L<R, Defining variables sum, if sum[i] + sum[L] + sum[R] = 0 Continue to cycle .

（5） if sum<0, The left pointer moves back ,sum > 0 The right pointer moves forward

（6） Be careful , if L and R The next element of is equal , To get rid of duplicate ！

（7） call sort() Sort 、aList() Replace the array with a set .

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        int l = nums.length;
        // Special judgement 
        if(nums ==  null &&  l< 3) return list;
        // Sort 
        Arrays.sort(nums);
        for(int i = 0; i < l ;i++){
           if(i > 0 && nums[i] == nums[i - 1 ]) continue;
            int L = i + 1;// Left pointer 
            int R = l - 1;// Right pointer 
            while( L <  R){
                int sum  = nums[i] + nums[L] + nums[R];
                if(sum == 0){
                //Arrays.asList() Convert an array to a collection 
                    list.add(Arrays.asList(nums[i],nums[L],nums[R]));
                    while(L < R && nums[L] == nums[L + 1])    L ++;// If the value of the next element of the left pointer is equal to the current value , There will be repeated results , Duplicate values need to be removed 
                    while(L < R && nums[R] == nums[R - 1])    R--;
                    // If the value of the next element of the right pointer is equal to the current value , There will be repeated results , Duplicate values need to be removed 
                    L++;
                    R--;
                }
                else if(sum < 0) L++;
                else if(sum > 0) R--;
            }
        }
        return list;

    }
}