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

利用map来解题。

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        Map<Integer,Integer> map = new HashMap<>();
        if(A == null || B == null || C == null || D == null) {
            return 0;
        }
        for(int i = 0;i < A.length;i++) {
            for(int j = 0;j < B.length;j++) {
                if(map.containsKey(A[i] + B[j])) {
                    map.put(A[i] + B[j],map.get(A[i] + B[j]) + 1);
                } else {
                    map.put(A[i] + B[j],1);
                }
            }
        }
        int ans = 0;
        for(int i = 0;i < C.length;i++) {
            for(int j = 0;j < D.length;j++) {
                int num = -(C[i] + D[j]);
                if(map.containsKey(num)) {
                    ans += map.get(num);
                }
            }
        }
        
        return ans;
    }
}

leetcode上大佬的代码

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        Map<Integer, Integer> ab = new HashMap<>();
        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) {
                res += ab.getOrDefault(-(c+d), 0);
            }
        }
        
        return res;
    }
}

不得不赞叹，代码竟然如此精简。。。

其中的getOrDefault方法在java8中的源码如下：

default V getOrDefault(Object key, V defaultValue) {

       V v;        

       return (((v = get(key)) != null) || containsKey(key)) ? v  : defaultValue;

   }

意思就是当Map集合中有这个key时，就使用这个key值，如果没有就使用默认值defaultValue