标签:should ... OLE 超时 code develop extends latest evel
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.
直接遍历即可,时间复杂度O(n)。但是超时了,所以不写了
二分查找法
/* 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 low = 1;
int high = n;
while(low < high) {
int mid = low + (high - low) / 2;
if(isBadVersion(mid)) high = mid;
else low = mid + 1;
}
return high;
}
}
[leetcode]278.First Bad Version
标签:should ... OLE 超时 code develop extends latest evel
原文地址:https://www.cnblogs.com/shinjia/p/9802095.html