LeetCode 278

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.
ince 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.

 

 1 /*************************************************************************
 2     > File Name: LeetCode278.c
 3     > Author: Juntaran
 4     > Mail: JuntaranMail@gmail.com
 5     > Created Time: Thu 19 May 2016 20:01:47 PM CST
 6  ************************************************************************/
 7 
 8 /*************************************************************************
 9     
10     First Bad Version
11     
12     You are a product manager and currently leading a team 
13     to develop a new product. Unfortunately, the latest version of 
14     your product fails the quality check. 
15     ince each version is developed based on the previous version, 
16     all the versions after a bad version are also bad.
17 
18     Suppose you have n versions [1, 2, ..., n] and you want to find out 
19     the first bad one, which causes all the following ones to be bad.
20 
21     You are given an API bool isBadVersion(version) 
22     which will return whether version is bad. 
23     Implement a function to find the first bad version. 
24     You should minimize the number of calls to the API.
25 
26  ************************************************************************/
27 
28 #include "stdio.h"
29 
30 // Forward declaration of isBadVersion API.
31 bool isBadVersion(int version);
32 
33 int firstBadVersion(int n) 
34 {
35     int low = 1;
36     int high = n;
37     int middle;
38     
39     while( low <= high )
40     {
41         middle = low + ( high - low ) / 2;        //此处一定要用low+(high-low)/2 如果使用(low+high)/2 可能溢出
42         if( isBadVersion(middle) == true )
43         {
44             high = middle - 1;
45         }
46         else
47         {
48             low = middle + 1;
49         }
50     }
51     return low;
52 }

 

posted @ 2016-05-20 12:54  Juntaran  阅读(159)  评论(0)  编辑  收藏  举报