Ideas

Train of thought ： Brute force .

Sum arrays in pairs . But it will time out .

class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        int[] res1 = sum(nums1,nums2);
        int[] res2 = sum(nums3,nums4);
        int l1 = res1.length;
        int l2 = res2.length;
        int res=0;
        for(int i =0 ;i<l1;i++){
            for(int j = 0;j<l2;j++){
                if(res1[i]+res2[j] == 0)
                    res++;
            }
        }
        return res;
    }
    public int[] sum(int[] nums1, int[] nums2){
        int l1 = nums1.length;
        int l2 = nums2.length;
        int[] res = new int[l1*l2];
        int index = 0;
        for(int i =0 ;i<l1;i++){
            for(int j = 0;j<l2;j++){
                res[index] = nums1[i]+nums2[j];
                index++;
            }
        }
        return res;
    }
}

Train of thought two ： Using hash table

The hash table can save the repeated results of the sum in two arrays in the hash table , It can reduce the time complexity .

Use the sum of the elements in the first two arrays map Storage ,map Of key Is and , The value is 1, If key Repeat will value++;

0- The sum of the last two arrays is the sum of the first two arrays , That is to say map Of key, Take the corresponding value That's all right. .

Code

class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        Map<Integer,Integer>map = new HashMap<>();
        int res = 0;
        for(int i =0;i<nums1.length;i++){
            for(int j =0;j<nums2.length;j++){
                int temp = nums1[i]+nums2[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) {
                int temp = i + j;
                if (map.containsKey(0 - temp)) {
                    res += map.get(0 - temp);
                }
            }
        }
        return res;
    }
}