LeetCode - 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:
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

这道题借鉴了2sum的解法， 创建一个hashmap， 先遍历A,B所有的组合把相反的和和出现次数存进去， 在遍历C,D的所有组合，找匹配。

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        Map<Integer, Integer> map =new HashMap<>();
        int count = 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(0-sum, map.getOrDefault(0-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];
                if(map.containsKey(sum)){
                    count += map.get(sum);
                }
            }
        }
        return count;
        
    }
}