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:
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

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

比4sum简单，因为不用考虑到重复元素（元素重复但是下标不重复）

解法很巧妙，用hashmap存放前两个数组每个pair和出现的次数，

然后在第三第四个数组中找有无一样的值，如果有就加上该次数，因为次数是AB数组独特的所以可以直接加。