LeetCode 454. Add four numbers II

Title Description ： Here are four integer arrays nums1、nums2、nums3 and nums4 , The length of the array is n , Please calculate how many tuples there are (i, j, k, l) To meet the ：

0 <= i, j, k, l < n
nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

Example 1：

 Input ：nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
 Output ：2
 explain ：
 Two tuples are as follows ：
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0

Example 2：

 Input ：nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
 Output ：1

Tips ：

n == nums1.length
n == nums2.length
n == nums3.length
n == nums4.length
1 <= n <= 200
-228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228

Thinking analysis ：

1. First define One map aggregate ,key discharge a and b Sum of two numbers ,value discharge a and b The number of times the sum of two numbers appears .
2. Traverse num1 and nums2 Array , Count the sum of two array elements , And the number of times , Put it in map in .
3. Definition int Variable res, Used for statistical a+b+c+d = 0 Number of occurrences .
4. In a traverse nums3 and nums4 Array , Find out if 0-(c+d) stay map If you've seen it in the past , Just use res hold map in key Corresponding value That is, the number of occurrences .
5. Finally, the statistics are returned res That's all right.

Code implementation ：

 public static int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4){
    
        Map<Integer, Integer> map = new HashMap<>();
        int temp;
        int res = 0;
        //  Count the sum of the elements in two arrays , At the same time, count the number of occurrences ,  Put in map
        for(int i : nums1){
    
            for(int j : nums2){
    
                temp = i + j;
                if(map.containsKey(temp)){
    
                    map.put(temp, map.get(temp) + 1);
                }else {
    
                    map.put(temp, 1);
                }
            }
        }

        //  Count the sum of the remaining two elements , stay map Find out if there is an addition to 0 The situation of , Record the number of times at the same time 
        for(int i : nums3){
    
            for(int j : nums4){
    
                temp = i + j;
                if(map.containsKey(0 - temp)){
    
                   res += map.get(0-temp)
                }
            }
        }
        return res;
    }

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/4sum-ii
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