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:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (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;    }}