原题链接在这里：

题目：

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

题解：

考虑这题时, 4个list太复杂，应该往简化版考虑. 如果只有2个list, 就很清楚了，一个list的值放进HashMap<Integer, Integer> hm中计算key的count.

再iterate 另一个list, 看hm中是否有当前数字的负数, 若有的话，有几个就有几种组合.

再反过来考虑4个list, 其实就是两两合并. AB合并, CD合并 与 AC合并, BD合并其实没有区别, 从结果的角度考虑，其实是没有变化的.

Time Complexity: O(n^2). n = A.length.

Space: O(n).

AC Java:

1 public class Solution { 2     public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { 3         int res = 0; 4         int len = A.length; 5         HashMap
     
       hm = new HashMap
      
       (); 6          7         for(int i = 0; i
      
     

与类似.