LeetCode 1248. Statistics "beautiful sub-array" (to be reviewed)

1. Title

Gives you an array of integers nums and an integer k.

If a continuous subarray exactly k in odd numbers, we believe that this is a sub-array "Yumiko array."

Please return the number of "beautiful subarrays" in this array.

示例 1：
输入：nums = [1,1,2,1,1], k = 3
输出：2
解释：包含 3 个奇数的子数组是 [1,1,2,1] 和 [1,2,1,1] 。

示例 2：
输入：nums = [2,4,6], k = 1
输出：0
解释：数列中不包含任何奇数，所以不存在优美子数组。

示例 3：
输入：nums = [2,2,2,1,2,2,1,2,2,2], k = 2
输出：16
 
提示：
1 <= nums.length <= 50000
1 <= nums[i] <= 10^5
1 <= k <= nums.length

Source: LeetCode (LeetCode)
link: https://leetcode-cn.com/problems/count-number-of-nice-subarrays The
copyright belongs to the deduction network. Please contact the official authorization for commercial reprint, and please indicate the source for non-commercial reprint.

2. Problem solving

2.1 Record odd pos

• Find the even number of k odd numbers, multiply it is the number of solutions

class Solution {
public:
    int numberOfSubarrays(vector<int>& nums, int k) {
    	int i, cnt = 0, n = nums.size(), count = 0;
    	vector<int> oddPos(n+2);
    	for(i = 0; i < n; ++i)
    	{
    		if(nums[i] & 1)//奇数
    			oddPos[++cnt] = i;
    	}
    	oddPos[0] = -1, oddPos[++cnt] = n;//边界，假设两边有0个偶数
    	for(i = 1; i+k <= cnt; ++i)
    		count += (oddPos[i]-oddPos[i-1])*(oddPos[i+k]-oddPos[i+k-1]);
	    return count;
    }
};

372 ms 66.1 MB

2.2 Prefix and

class Solution {
public:
    int numberOfSubarrays(vector<int>& nums, int k) {
    	int i, oddcnt = 0, n = nums.size(), count = 0;
    	vector<int> preOddCnt(n+1,0);
    	preOddCnt[0] = 1;//边界
    	for(i = 0; i < n; ++i)
    	{
    		oddcnt += (nums[i]&1);//奇数多少个了
            preOddCnt[oddcnt] += 1;//这么多个奇数的数组 +1 个
            if(oddcnt >= k)//奇数够个数了
    		    count += preOddCnt[oddcnt-k];
    		    //以i结束的长度为k个奇数的数组个数 preOddCnt[oddcnt-k]
    	}
	    return count;
    }
};

368 ms 65.8 MB