LeetCode-974. Subarray Sums Divisible by K

1、题目

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

2、思路

这道题要求原数组所构成的连续子数组元素和能被K整除的所有子数组个数，我们定义sum(i, j)表示下标从i到j-1的子数组元素的和，则sum(i, j) = sum(0, j) - sum(0, i), 所以我们知道了sum(0, a)(1<=a<=N N为数组元素个数)，任何一个子数组的和我们都可以利用sum(i, j)=sum(0, j)-sum(0, i)求出来.于是有以下推导

任意子数组和sum(i, j)能被K整除，即sum(i, j) % K = 0 ==> [sum(0, j)-sum(0, i)] % K = 0 ==> sum(0, j) % K = sum(0, i) % K,

于是我们只需要知道sum(0, a)中哪些和存在同余的关系，即sum(0, x)与sum(0, y)同余，关于同余的概念可以去数论中了解。

举例：A = [4, 5, 0, -2, -3, 1]

1). 我们首先计算sum(0, a)存到数组中得 P=[0, 4, 9, 9, 7, 4, 5]

2). 然后按照同余关系进行计数得:

与0同余的有(0, 5) => (P[0], P[6])

与2同余的有(7) =>(P[4])

与4同余的有(4, 9, 9, 4) =>(P[1], P[2], P[3], P[5])

3).对于同余的P[i]中，两两组合所得到的组合数相加就是要求的结果

3 实现

public class Solution {
    public int subarraysDivByK(int[] A, int K) {
        int n = A.length;
        int[] P = new int[n+1];
        for (int i = 0; i < n; i++){
            P[i+1] = P[i] + A[i];
        }

        int[] C = new int[K];
        for (int x : P){
            C[(x % K + K) % K]++;
        }

        int ans = 0;
        for (int c : C){
            ans += c * (c-1) / 2;
        }
        return ans;
    }
}

LeetCode-974. Subarray Sums Divisible by K