【leetcode】278. First Bad Version（Python & C++）

278. First Bad Version

题目链接

278.1 题目描述：

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.

278.2 解题思路：

思路一：二分查找法，不做赘述。

思路二：同一，写法不同。

278.3 C++代码：

1、思路一代码（0ms）：

class Solution132_2 {
public:
int firstBadVersion(int n) {
int low = 0;
int high = n;
int mid;
while (high-low>1)
{
mid = low + (high - low) / 2;
if (isBadVersion(mid))
{
high = mid;
}
else
{
low = mid;
}
}
return high;
}
};

2、思路二代码（0ms）

class Solution132_3 {
public:
int firstBadVersion(int n) {
int low = 1;
int high = n;
int mid;
while (high>low)
{
mid = low + (high - low) / 2;
if (isBadVersion(mid))
{
high = mid;
}
else
{
low = mid + 1;
}
}
return high;
}
};

278.4 Python代码：

1、思路一代码（28ms）

class Solution(object):
def firstBadVersion(self, n):
"""
:type n: int
:rtype: int
"""
low=0
high=n
mid=0
while high-low>1:
mid=low+(high-low)/2
if isBadVersion(mid)==True:
high=mid
else:
low=mid
return high

2、思路二代码（29ms）

class Solution1(object):
def firstBadVersion(self, n):
"""
:type n: int
:rtype: int
"""
low=1
high=n
mid=0
while low<high:
mid=low+(high-low)/2
if isBadVersion(mid)==True:
high=mid
else:
low=mid+1
return high