[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))
[Lintcode]74. First Bad Version/[Leetcode]278. First Bad Version
原文地址:https://www.cnblogs.com/siriusli/p/10357078.html
时间: 2024-10-08 22:26:37