[LeetCode/LintCode] First Bad Version

Problem

The code base version is an integer start from 1 to n. One day, someone committed a bad version in the code case, so it caused this version and the following versions are all failed in the unit tests. Find the first bad version.

You can call isBadVersion to help you determine which version is the first bad one. The details interface can be found in the code's annotation part.

Notice

Please read the annotation in code area to get the correct way to call isBadVersion in different language. For example, Java is SVNRepo.isBadVersion(v)

Example

Given n = 5:

isBadVersion(3) -> false
isBadVersion(5) -> true
isBadVersion(4) -> true
Here we are 100% sure that the 4th version is the first bad version.

Challenge

You should call isBadVersion as few as possible.

Solution

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int start = 1, end = n;
        while (start <= end) {
            int mid = start+(end-start)/2; 
            //分析最后一次循环的时刻：
            //当start与end相差小于2时，mid总是start, 那么如果start是bad，下一次直接跳出循环，返回start
            //当start与end相差大于2时，mid是bad，end变成mid-1，如果mid是first bad，循环结束的条件将是：
            //end stays on the same position, start = end+1 --> start goes to the first bad position
            //循环结束，返回start
            if (isBadVersion(mid)) {
                end = mid-1;
            } else {
                start = mid+1;
            }
        }
        return start;
    }
}

Update 2018-9

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

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        if (n == 0) return 0;
        int i = 1, j = n;
        while (i < j) {
            int mid = i+(j-i)/2;
            //mid could be the first bad, so DON'T do: j = mid-1
            if (isBadVersion(mid)) j = mid;
            //when mid is not bad, just DO: i = mid+1 
            else i = mid+1;
        }
        return j;
    }
}