Hold住Leetcode——First Bad Version

bool isBadVersion(int version);
class Solution {public:    int firstBadVersion(int n) {        int front=1,end=n;        if(isBadVersion(1))        return 1;                while(1)        {        int mid=(end-front)/2+front;        bool TF=isBadVersion(mid);        if(TF==1)        end=mid-1;        else        {front=mid+1;     //先找到一个good值，然后判断其后面一个是不是bad，如果是就返回mid+1        if(isBadVersion(mid+1))//         return mid+1;        }        }            }};

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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题目的大概意思是，找出一系列生产步骤中开始出错的那一步，已知从第一步出错开始，就会一步错，步步错。

解题的主要思想仍然是二分法查找。