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
思路: 类似560的题,因为 sum[j] - sum[i] %K == 0等价于
sum[j]%K == sum[i]% 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;
count += map[sum];
map[sum]++;
}
return count;
}
}