原题
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。
示例:
输入: A = [4,5,0,-2,-3,1], K = 5
输出: 7
解释:
有 7 个子数组满足其元素之和可被 K = 5 整除:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
提示:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
解法
思想
看到连续的子数组的和就要想到前缀和
代码
class Solution {
public int subarraysDivByK(int[] A, int K) {
int[] remainder = new int[K];
remainder[0] = 1;
int sum = 0;
int count = 0;
for(int i = 0;i<A.length;i++){
sum += A[i];
int remain = sum%K;
if(remain<0) remain += K;
count += remainder[remain];
remainder[remain]++;
}
return count;
}
}
原创文章,作者:彭晨涛,如若转载,请注明出处:https://www.codetool.top/article/leetcode974-%e5%92%8c%e5%8f%af%e8%a2%abk%e6%95%b4%e9%99%a4%e7%9a%84%e5%ad%90%e6%95%b0%e7%bb%84/