6/5 每日一題 (二分法)

 You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

 

Example 1:

Input: n = 5, bad = 4
Output: 4
Explanation:
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
Then 4 is the first bad version.

Example 2:

Input: n = 1, bad = 1
Output: 1

 

Constraints:

• 1 <= bad <= n <= 231 - 1

# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:

class Solution:
    def firstBadVersion(self, n: int) -> int:
        '''使用二分法'''
        left=1
        right=n
        while left < right: 
            mid = left + ((right-left)//2) 
            #這邊要記得 不能是單純的左右端點相加/2
            #mid必定要在left的右邊所以要加上left
            #確保mid在左端點右邊
            #[1,2,3,4,5], 壞的是[4,5]
            if isBadVersion(mid): 
                #True就表示在mid還再壞的範圍
                #右指針往左移動
                right=mid
            else:
                #表示mid介於與上一個之間,移動左指針確認
                left=mid+1
        return left
        

-------參考解答

# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:

class Solution:
    def firstBadVersion(self, n: int) -> int:
        left,right=0,n-1
        while left<=right:
            mid=(left+right)//2
            if isBadVersion(mid)==False:
                left=mid+1
            else:
                right=mid-1
        return left