leetcode First Bad Version（二分查找）

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.

这道题比较简单，可是我却也琢磨了很久，其实看到这道题第一个反应就是二分查找。专业点说，这是一个寻找左边界的问题。

思路就是：如果头部是bad version的话，可以返回；如果mid是坏的话，坏的源头必然在head和mid之间，那么尾部就可以变为mid了。如果mid不是坏的，那么坏的必然在mid和尾部之间，所以头部会变成mid。（就是二分查找了！）

但是刚开始mid=(head+tail)/2会time exceed（因为直接加head+tail有可能会溢出）。说明【自己没有良好的代码习惯】，以前看书的时候就说，二分查找，最好写成mid=low+(high-low)/2。因为time exceed，又考虑了别的思路，导致这道题做了很久。

// Forward declaration of isBadVersion API.

bool isBadVersion(int version);

class Solution {

public:

    int firstBadVersion(int n) {

        if(n==) return -;

        int head=,tail=n,mid=;

        while(head<=tail){

            mid=head+(tail-head)/;

            if(isBadVersion(head)) return head;

            if(isBadVersion(mid)) {head++; tail=mid;}

            else head=mid+;

        }

        return -;

    }

};

巴特西

leetcode First Bad Version（二分查找）

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