4Sum II

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

tips：kSum问题中一个子问题，时间复杂度可以降到O(2).详细可以参考http://www.mamicode.com/info-detail-616720.html

public class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int n = A.length;
        if( n == 0 ) return 0;
        HashMap<Integer,Integer> ab = new HashMap<Integer,Integer>();
        for(int a:A){
            for(int b:B){
                ab.put(a+b,ab.getOrDefault(a+b,0) + 1);
            }
        }
        int res = 0;
        for(int c:C){
            for(int d:D){
                int part2 = c + d;
                int part1 = - part2;
                res += ab.getOrDefault(part1,0);
            }
        }
        return res;



    }
}