278. First Bad Versionbinary search

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 will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Example:

Given n = 5,and version = 4 is the first bad version.

call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true

Then 4 is the first bad version.

# The isBadVersion API is already defined for you.
# @param version,an integer
# @return a bool
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self,n):
        """
        :type n: int
        :rtype: int
        """
        left,right = 1,n
        while left<right:
            mid = (left+right)//2
            if not isBadVersion(mid):
                left = mid+1
            else:
                right = mid
        return mid

这道题不方便本地调试 这道题直接搜索会TLE，只能用二分搜索。 为什么left = mid + 1而right = mid呢？因为你可以确定包括left的左边不会是答案所在，而right那个点不能确定是否是答案所在。 注意//是整数除法