[Lintcode]74. First Bad Version/[Leetcode]278. First Bad Version

Posted 2021-02-08 siriusli

[Lintcode]74. First Bad Version

[Leetcode]278. First Bad Version

• 本题难度: [Lintcode]Medium/[Leetcode]

• Topic: Binary Search

  Description

我的代码 for Leetcode(在Lintcode中因为isBadVersion调用太多次无法通过)

# 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
        """
        if isBadVersion(1) is True:
            return 1
        else:
            if n==1:
                return 0
        l = 2
        r = n
        m = (l+r)//2
        while(l<=r):
            print(l,m,r)
            t1 = isBadVersion(m-1)
            t2 = isBadVersion(m)
            print(t1,t2)
            if t1 is False and t2 is True:
                print("ok")
                return m
            else:
                if t2 is False:
                    l = m+1
                else:
                    if t1 is True:
                        r = m-1
            m = (l+r)//2

别人的代码 for Lintcode

#class SVNRepo:
#    @classmethod
#    def isBadVersion(cls, id)
#        # Run unit tests to check whether verison `id` is a bad version
#        # return true if unit tests passed else false.
# You can use SVNRepo.isBadVersion(10) to check whether version 10 is a 
# bad version.
class Solution:
    """
    @param n: An integer
    @return: An integer which is the first bad version.
    """
    def findFirstBadVersion(self, n):
        # write your code here

        r = n-1
        l = 0
        while(l<=r):
            mid = l + (r-l)//2
            if SVNRepo.isBadVersion(mid)==False:
                l = mid+1
            else:
                r = mid-1
        return l

思路

大思路是二分法，但是我想岔了。
其实无需找到一个自身为true，前一个为false的id，只需要找到最后一个为false的id即可。这样可以减少一次比较。

• 时间复杂度: O(log(n))