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The first wrong version of day036 (binary search method) | primary algorithm | Python

編輯：Python

Make a little progress every day , It's already great , Insist on , Don't be too tired , Refuse to roll in , Start with practice every day , Ten minutes a day , Live happily all your life ！ The epidemic is still repeated , Everyone wear masks ~ Continue to continue , Come on , Today and Brother cheshen Work together to improve your Python Programming and Interview ability Well , brush ladder on high buildings ~

Put on what I took Photo Well ！

Recommend a song every day ： fragrance of rice ——Jay Chou

The following is my Ladder integral rule ：

At least one question a day ： An integral +10 branch
If you do one more question （ Or one more way to answer ）, Then the points of the day +20 branch （+10+10）
If you do more than three , Start with question 3 +20 branch （ Such as ： If you do three questions, you will score -10+10+20=40; If you do four questions, you will score –10+10+20+20=60）
Initial classification 100 branch
If you haven't done the problem for one day , Then deduct points -10 branch （ Saturday 、 Except Sunday ： rest ）
insist ！！！

Primary algorithm

List of questions

Linked list

stem

You are the product manager , Currently leading a team to develop new products . Unfortunately , The latest version of your product doesn't pass the quality test . Because each version is based on the previous version , So all versions after the wrong version are wrong .

Suppose you have n A version [1, 2, …, n], You want to find out the first wrong version that caused all subsequent versions to go wrong .

You can call bool isBadVersion(version) Interface to determine version number version If there is an error in unit test . Implement a function to find the first wrong version . You should try to minimize calls to API The number of times .

Example 1：

Input ：n = 5, bad = 4
Output ：4
explain ：
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
therefore ,4 It's the first wrong version .

Example 2：

Input ：n = 1, bad = 1
Output ：1

Dichotomy search

analysis ：

Is it a little silly to see this topic , It's ridiculous .
But it doesn't seem very difficult to look at it , We use dichotomy to reduce the number of test interface calls , Median . Constant temptation , Judge whether it is False, Then keep narrowing the range , Finally find the version number .

# The isBadVersion API is already defined for you.# @param version, an integer# @return an integer# def isBadVersion(version):class Solution: def firstBadVersion(self, n): """ :type n: int :rtype: int """ # Binary search  st = 0 end = n while st <= end: mid = st + (end - st)//2 if isBadVersion(mid): end = mid - 1 else: st = mid + 1 return st

It seems that there is only such an answer , Two days and two questions , Finish sorting and searching .