【leetcode】974. Subarray Sums Divisible by K

题目如下：

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.
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

解题思路：本题需要用到一个数学规律，如果a%c = b%c，那么(a-b)%c=0。我的解法就是从后往前遍历数组，依次累加每个元素的值并记为sum，同时用字典保存sum%K作为key值出现的次数。同时每累加一个元素，只要去字典中查找历史sum%K出现的次数，这个次数就是从以这个元素作为起点满足条件的子数组的个数。特别注意的是，如果sum%K=0，那么表示这个元素本身就满足条件，次数要+1。

代码如下：

class Solution(object):def subarraysDivByK(self, A, K):""":type A: List[int]:type K: int:rtype: int"""dic = {}count = 0res = 0for i in A[::-1]:count += iif count%K in dic:res += dic[count%K]if count % K == 0:res += 1dic[count%K] = dic.setdefault(count%K,0)+1return res

转载于:https://www.cnblogs.com/seyjs/p/10317990.html

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