Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.Example:Input:A = [ 1, 2]B = [-2,-1]C = [-1, 2]D = [ 0, 2]Output:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

这道题借鉴了2sum的解法， 创建一个hashmap， 先遍历A,B所有的组合把相反的和和出现次数存进去， 在遍历C,D的所有组合，找匹配。

class Solution {    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {        Map
     
       map =new HashMap<>();        int count = 0;                for(int i = 0; i < A.length; i++){            for(int j = 0; j< B.length; j++){                int sum = A[i] + B[j];                map.put(0-sum, map.getOrDefault(0-sum,0)+1);            }        }                for(int i = 0; i