Leetcode NO.278 First Bad Version

本题题目要求如下：

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.
本题就是divide and conquer，跟binary search很像，就是做了一点改变：如果验证该位为bad version，还需要检查前一位是不是，如果是，就返回，如果不是，还得继续divide & conquer，另外需要注意的是要避免integer溢出，解决方法是用：
int mid = low + (high - low) / 2;而不是用
int mid = (low + high) / 2;
然后本题的完整代码如下所示：

// Forward declaration of isBadVersion API.
bool isBadVersion(int version);

class Solution {
public:
int firstBadVersion(int n) {
int low = 1;
int high = n;
int mid = low + (high - low) / 2;
while (low <= high) {
if (!isBadVersion(mid)) {
low = mid + 1;
}
else {
if (!isBadVersion(mid-1)) {
return mid;
}
else {
high = mid - 1;
}
}
mid = low + (high - low) / 2;
}
}
};