LeetCode 454. 4Sum II

原题链接在这里：https://leetcode.com/problems/4sum-ii/?tab=Description

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:
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

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:

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

与4Sum类似.