Open azl397985856 opened 1 month ago
class Solution:
def firstBadVersion(self, n: int) -> int:
left = 1
right = n
while left <= right:
mid = left + (right - left)//2
if isBadVersion(mid):
right = mid - 1
else:
left = mid + 1
return left
Time: O(LogN)
Space:O(1)
class Solution:
def firstBadVersion(self, n: int) -> int:
if n == 1: return 1
left,right = 1, n
while True:
if left > right:
break
mid = (left + right) // 2
if not isBadVersion(mid):
left = mid + 1
else: right = mid - 1
return left
复杂度分析
时间复杂度:O(logN) 空间复杂度:O(1)
class Solution:
def firstBadVersion(self, n: int) -> int:
l, r = 1, n
while l <= r:
mid = (l + r) // 2
if isBadVersion(mid): r = mid - 1
else: l = mid + 1
return l
# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:
class Solution:
def firstBadVersion(self, n: int) -> int:
left = 1
right = n
while left < right:
mid = left + (right - left) // 2
if isBadVersion(mid):
right = mid
else:
left = mid + 1
return left
using brute force will get TLE Binary search
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */
public class Solution extends VersionControl {
public int firstBadVersion(int n) {
int left =1;
int right = n;
while(left < right){
//making mid
int mid = left + (right - left)/2;
if (isBadVersion(mid))
right = mid;
// narrow down the searching area
else
left = mid +1;
}
return right;
}
}
278. 第一个错误的版本
入选理由
暂无
题目地址
https://leetcode-cn.com/problems/first-bad-version
前置知识
题目描述
假设你有 n 个版本 [1, 2, ..., n],你想找出导致之后所有版本出错的第一个错误的版本。
你可以通过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错。实现一个函数来查找第一个错误的版本。你应该尽量减少对调用 API 的次数。
示例:
给定 n = 5,并且 version = 4 是第一个错误的版本。
调用 isBadVersion(3) -> false 调用 isBadVersion(5) -> true 调用 isBadVersion(4) -> true
所以,4 是第一个错误的版本。