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974.Subarray-Sums-Divisible-by-K

974. Subarray Sums Divisible by K

题目地址

题目描述

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.
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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]
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Note:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000

代码

Approach #1

Time Complexity: O(N) Space Complexity: O(K)
if sum[0, i] % K == sum[0, j] % K, sum[i + 1, j] is divisible by by K.
class Solution {
public int subarraysDivByK(int[] A, int K) {
int[] map = new int[K];
map[0] = 1;
int count = 0, sum = 0;
for(int a : A) {
sum = (sum + a) % K;
if(sum < 0) sum += K; // Because -1 % 5 = -1, but we need the positive mod 4
count += map[sum];
map[sum]++;
}
return count;
}
}