LeetCode 454. 4Sum II

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题目要求

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

题意解析

存在四个数列,从各个数列中取1个数出来,当这些数的和为0是算一组,求一共有多少组这样的数。

解法分析

这道题充分利用HashMap,只需要O(n2)的时间复杂度即可。

解题代码

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Map<Integer, Integer> abSum = new HashMap<>();
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
if (A == null || A.length <= 0 || B == null || B.length <= 0
|| C == null || C.length <= 0 || C == null || C.length <= 0) {
return 0;
}
int number = 0;
for (int i = 0; i < A.length; i++) {
for (int j = 0; j < B.length; j++) {
int tmp = A[i] + B[j];
abSum.put(tmp, abSum.getOrDefault(tmp, 0) + 1);
}
}
for (int i = 0; i < C.length; i++) {
for (int j = 0; j < D.length; j++) {
number += abSum.getOrDefault(-(C[i] + D[j]), 0);
}
}
return number;
}