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:

  1. Input:
  2. A = [ 1, 2]
  3. B = [-2,-1]
  4. C = [-1, 2]
  5. D = [ 0, 2]
  7. Output:
  8. 2
  10. Explanation:
  11. The two tuples are:
  12. 1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
  13. 2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

题意

给定四个数组，在各个数组中取出一个数使其和为0，求这样的数对的个数。

思路

用HashMap记录其中两个数组的组合情况，再遍历另外两个数组的组合进行处理。

代码实现

Java

  1. class Solution {
  2.     public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
  3.         int count = 0;
  4.         Map<Integer, Integer> hash = new HashMap<>();
  6.         for (int i = 0; i < A.length; i++) {
  7.             for (int j = 0; j < B.length; j++) {
  8.                 int sum = A[i] + B[j];
  9.                 hash.put(sum, hash.getOrDefault(sum, 0) + 1);
  10.             }
  11.         }
  13.         for (int i = 0; i < C.length; i++) {
  14.             for (int j = 0; j < D.length; j++) {
  15.                 int target = - C[i] - D[j];
  16.                 if (hash.containsKey(target)) {
  17.                     count += hash.get(target);
  18.                 }
  19.             }
  20.         }
  22.         return count;
  23.     }
  24. }