Description
Given an array A
of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K
.
给定一个整数数组 A
,返回其中元素之和可被 K
整除的(连续、非空)子数组的数目。
题目链接:https://leetcode.com/problems/subarray-sums-divisible-by-k/
Difficulty: medium
Example 1:
Input: A = [4,5,0,-2,-3,1], K = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by K = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
Note:
- 1 <= A.length <= 30000
- -10000 <= A[i] <= 10000
- 2 <= K <= 10000
分析
- updating(Solution)
参考代码
class Solution(object):
def subarraysDivByK(self, A, K):
P = [0]
for x in A:
P.append((P[-1] + x) % K)
count = collections.Counter(P)
return sum(v*(v-1)/2 for v in count.values())