• Total Accepted: 8398
• Total Submissions: 18801
• Difficulty: Medium
• Contributors: Samuri

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

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【题目分析】

给定四个长度相同的数组，在每个数组中取一个数字，在所有的组合中和为零的组合有多少个？

【思路】

把四个数组分为两组，每组包含两个数组。把其中一组中的任意两个值和存入hashmap中，然后在hashmap查找另外两个数组的值的组合。这其实是相当于转化为了一个two sum问题。

【java代码】

 public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {

     Map<Integer, Integer> map = new HashMap<>();

     for(int i=0; i<C.length; i++) {

         for(int j=0; j<D.length; j++) {

             int sum = C[i] + D[j];

             map.put(sum, map.getOrDefault(sum, 0) + 1);

         }

     }

     int res=0;

     for(int i=0; i<A.length; i++) {

         for(int j=0; j<B.length; j++) {

             res += map.getOrDefault(-1 * (A[i]+B[j]), 0);

         }

     }

     return res;

 }

代码中的map.getOrDefault(sum, 0)相比先在map中查找再取数的操作是比较高效的。