> For the complete documentation index, see [llms.txt](https://rachel2011.gitbook.io/leetcode_cc150/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://rachel2011.gitbook.io/leetcode_cc150/binary-search/first-bad-version.md).

# First Bad Version

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.

题目大意：

你是一名产品经理，正在领导团队开发一个新产品。不幸的是，产品的最新版本没有通过质量检测。由于每一个版本都是基于上一个版本开发的，某一个损坏的版本之后的所有版本全都是坏的。

假设有`n`个版本`[1, 2, ..., n]`，你想要找出第一个损坏的版本，它导致所有后面的版本都坏掉了。

给你一个API`bool isBadVersion(version)`，返回某一个版本是否损坏。实现一个函数找出第一个损坏的版本。你应该最小化调用API的次数。

### 解题思路：

**二分法（Binary Search），由于这题很有规律，好版本和坏版本一定有个边界，那么我们用二分法来找这个边界**，对mid值调用API函数，如果是坏版本，说明边界在左边，则把mid-1赋值给right，如果是好版本，则说明边界在右边，则把mid+1赋给left，最后返回left即可。需要注意的是，OJ里有个坑，那就是如果left和right都特别大的话，那么left+right可能会溢出，我们的处理方法就是变成left + (right - left) / 2，很好的避免的溢出问题。

为什么最后返回left？

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

class Solution(object):
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """
        left, right = 1, n
        while left <= right:
            mid = (left + right) / 2
            if isBadVersion(mid):
                right = mid - 1
            else:
                left = mid + 1
        return left
```

## 二刷：

这道题有点像在array中找出和target相等的最小index那道题

因为start＝mid，end＝mid不会错过badversion，所以最后循环结束，start和end中至少有一个badversion

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

class Solution(object):
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """

        start, end = 1, n
        while start+1 < end:
            mid = (start+end)/2
            if isBadVersion(mid):
                end = mid
            else:
                start = mid
        if isBadVersion(start):
            return start
        return end
```