Description

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: 2

Explanation: The two tuples are: 1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0 2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

解法

思路是把A和B中的数字两两求和，存入一个map，key是两个数的和，value是这个和出现的次数。 之后遍历C和D，取出-(C[i] + D[j])在出现的次数，累加起来即为结果。

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