4Sum II

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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

今天自己正八经写的代码估计也就3，4道。遇到过bfs遇到过tree，感觉还是基础不太扎实，什么时候用什么方法应该很清楚才是。

这道题是自己想出来的。想法比较intuitive, 先保存前两个数组的组合，然后再遍历后两个数组找是否有和的相反数存在在之前保存的hashmap中。

时间复杂度O(n2), 空间复杂度也是O(n2).

代码：

public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();        int count = 0;        for(int item1: A){            for(int item2: B){                int sum = item1+item2;                map.putIfAbsent(sum, 0);                map.put(sum, map.get(sum)+1);            }        }        for(int item3: C){            for(int item4 : D){                if(map.containsKey(-(item3+item4))){                    count+=map.get(-(item3+item4));                }            }        }        return count;    }

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