LeetCode1523- Count Odd Numbers in an Interval Range (Easy)

Given two non-negative integers low and high. Return the count of odd numbers between low and high (inclusive).

 

Example 1:

Input: low = 3, high = 7
Output: 3
Explanation: The odd numbers between 3 and 7 are [3,5,7].

Example 2:

Input: low = 8, high = 10
Output: 1
Explanation: The odd numbers between 8 and 10 are [9].

方法：Math
Time complexity: O(1) Space complexity: O(1)

class Solution:
    def countOdds(self, low: int, high: int) -> int:
        # low: odd && high: odd [3,9] 3, 5, 7,9 
        count = 0 
        if low % 2 == 1 and high %2 == 1:
            count = self.getNum(low, high)
        elif low % 2 == 1 and high % 2 == 0: #[3, 8]
            high -= 1 
            count = self.getNum(low, high)
        elif low %2 == 0 and high % 2 == 1: #[4, 7]
            low += 1 
            count = self.getNum(low, high) 
        else:
            low += 1 
            high -= 1 
            count = self.getNum(low, high) 
            
        return count 
    
    def getNum(self, low, high):
        count = (high - low -1) //2 + 2 
        return count