454. 4Sum II

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这里用hashmap，可以做到O(n^2)的复杂度，分别存储A和B各个元素的组合和，接下来用C和D各个组合和去查询这个hashmap有没有对应的负值。代码如下：
public class Solution {    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {        Map<Integer, Integer> map = new HashMap<Integer, Integer>();        int res = 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(sum, map.getOrDefault(sum, 0) + 1);            }        }        for (int i = 0; i < C.length; i ++) {            for (int j = 0; j < D.length; j ++) {                int sum = C[i] + D[j];                res += map.getOrDefault(-1 * sum, 0);            }        }        return res;    }}