Leetcode 974. Subarray Sums Divisible by K （前缀和数组典型题）

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该篇文章介绍了一种算法，用于解决给定整数数组nums和整数k，计算满足元素之和能被k整除的非空子数组数量。使用前缀和数组和哈希映射的方法，记录每个余数出现的频率来求解。

Subarray Sums Divisible by K
Medium

Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k.
A subarray is a contiguous part of an array.

Example 1:

Input: nums = [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]
Example 2:

Input: nums = [5], k = 9
Output: 0

Constraints:

1 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
2 <= k <= 104

解法1：前缀和数组+hashmap。
注意

i…j可以整除说明presums[i] % k == presums[j] %k. 所以我们保存<remainder, freq>的hashmap就可以。
remainder可以是负数。比如-2%5=-2。 3%5=3。但是它们的差是5，可以被5整除。所以 remainder = remainder < 0 ? remainder + k : remainder;
remainderFreq[0] = 1; 不然会漏一个。

class Solution {
public:
    int subarraysDivByK(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> presums(n + 1, 0);
        unordered_map<int, int> remainderFreq; //<remainder, freq>
        int res = 0;
        remainderFreq[0] = 1;
        for (int i = 1; i <= n; i++) {
            presums[i] = presums[i - 1] + nums[i - 1];
            int remainder = presums[i] % k;
            remainder = remainder < 0 ? remainder + k : remainder;
            if (remainderFreq.find(remainder) != remainderFreq.end()) {
                res += remainderFreq[remainder];
            }
            remainderFreq[remainder]++;
        }
        return res;
    }
};