【LeetCode & 剑指offer刷题】查找与排序题10：First Bad Version
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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.
Example:
Given n = 5, and version = 4 is the first bad version.
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
Then 4 is the first bad version.
// Forward declaration of isBadVersion API.
bool isBadVersion(int version);
//序列可看成（0,0,0,0,1,1,1,1,1,1,1,1,1,1）找第一个为1的数（可用lower_bound函数）
class Solution
{
public:
int firstBadVersion(int n) //版本序列为有序序列，可用二分查找
{
int left = 1,mid,right=n;
while(left<=right) //这里也可以写成<
{
mid = left + (right - left)/2; //防止（left+right）/2造成溢出风险
if(!isBadVersion(mid))
{
left = mid + 1; //为0时往右边找，left会移动到第一个为1的数的位置
}
else
{
right = mid - 1;//为1时往左边找，也可以写成right=mid
}
}
if(isBadVersion(left)) return left;
else return -1; //如果找不到就返回-1
}
};
/*
//一般的二分查找
while(left <= right) //注意这里为<=，因为要计算一次mid再返回
{
mid = left + (right - left)/2; //防止（left+right）/2造成溢出风险
if(a[mid]<key)
{
left = mid + 1;
}
else if(a[mid]>key)
{
right = mid -1;
}
else
{
return mid;
}
}
*/
