【LeetCode 1248】 Count Number of Nice Subarrays

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本文介绍了一种算法，用于计算给定整数数组中包含k个奇数的子数组数量。通过遍历数组并累计奇数个数，当达到k时，统计并更新计数，最终返回所有符合条件的子数组总数。

题目描述

Given an array of integers nums and an integer k. A subarray is called nice if there are k odd numbers on it.

Return the number of nice sub-arrays.

Example 1:

Input: nums = [1,1,2,1,1], k = 3
Output: 2
Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].

Example 2:

Input: nums = [2,4,6], k = 1
Output: 0
Explanation: There is no odd numbers in the array.

Example 3:

Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2
Output: 16

Constraints:

1 <= nums.length <= 50000
1 <= nums[i] <= 10^5
1 <= k <= nums.length

思路

累计当前区间内奇数的个数，等于k个时，左区间增加，统计个数。只有当右区间新覆盖了奇数时，重新开始当前计数。

代码

class Solution {
public:
    int numberOfSubarrays(vector<int>& nums, int k) {
        int n = nums.size();
        int res = 0, cnt = 0, cur = 0;
        for (int i=0, j=0; i<n; ++i) {
            if (nums[i] % 2) cnt++;
            if (cnt == k) cur = 0;
            while(cnt == k) {
                cur++;
                if (nums[j] % 2) cnt--;
                j++;
            }
            res += cur;
        }
        return res;
    }
};