题目：

给你两个非负整数 low 和 high 。请你返回 low 和 high 之间（包括二者）奇数的数目。

示例 1：

输入：low = 3, high = 7
输出：3
解释：3 到 7 之间奇数数字为 [3,5,7] 。

示例 2：

输入：low = 8, high = 10
输出：1
解释：8 到 10 之间奇数数字为 [9] 。

提示：

• 0 <= low <= high <= 10^9

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 第一种方法，直接判断奇数偶数，是奇数，计数器++；

class Solution {public int countOdds(int low, int high) {int cnt = 0;if(low%2!=0){while(low<=high){cnt++;low+=2;}}if(low%2==0){low++;while(low<=high){cnt++;low+=2;}}return cnt;}
}

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第二种，列出所有情况，high-low=0；low奇，high偶；high奇，low偶；low、high全奇全偶。

class Solution {public int countOdds(int low, int high) {int cnt = 0;if(high-low==0){if(low%2==0)return cnt;return cnt+1;}if(low%2==0&&high%2==0){cnt+=(high-low)/2;}else if(low%2!=0&&high%2!=0){cnt+=((high-low)/2+1);}else{cnt+=(high-low+1)/2;}return cnt;}
}

第三种，一行代码秒杀！

class Solution {public int countOdds(int low, int high) {return (high+1)/2 - low/2;}
}