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

一开始想了一个O（n^3）, space O(N)的做法，后来发现还可以优化

Solution： time O(N^2), space O(N^2)

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

         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;
     }
 }