#leetcode#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.

一个月没刷题leetcode又多了三十多道，walmartlabs工作差不多快一个月了，这次别把刷题放下了

老套路，二分查找

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int l = 1;
        int r = n;
        // while(l <= r){
        while(l < r){
            int m = l + (r - l) / 2;
            if(isBadVersion(m)){
                r = m;
            }else{
                l = m + 1;
            }
        }
        
        return r;
    }
}

这个模板我应该说比较熟了，但是第一次写的时候while循环的结束条件写成了 l <= r, 举个简单的例子， n等于2， first bad version是1，就过不了，

注意这里每一次对mid做判断之后，r ＝ m 或者 l ＝ m ＋ 1， r 有可能等于m，也就是说r 有可能等于 l， 因为 m >= l && m <= r, 这个时候再令结束条件为l 小于等于 r 的话显然就不对了