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

十分钟尝试

类似于之前的，放入map加1，第二次在map减去1，最后为0表示符合预期，对称的意思。

这个题目类似，sum等于0，意味着两个整数，两个负数。前两个数组元素任意相加的和存入map，后两个数组任意元素相加得到的和，从map取其负数，如果有则count加1，最后count就是所求

class Solution {
    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 count=0;
          for(int i=0;i<C.length;i++){
            for(int j=0;j<D.length;j++){
                int sum=C[i]+D[j];
                count+=map.getOrDefault(-1*sum,0);
            }
        }
        return count;
    }
}