454. 4Sum Ⅱ

题目描述

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

思路分析

这里四个数字的和为定值，可以参照两个数之和为定值的解法。4Sum = 2Sum + 2Sum。这里也是借助HashMap，先确定sum = A[i]+B[j]，存入hashMap中，然后对于C，D数组，判断HashMap中是否存在-（C[i]+D[j]），若存在取出对应的值。累加即为所有可能结果的个数。

代码

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