Leetcode:278. First Bad Version(JAVA)

最新推荐文章于 2022-07-28 16:41:37 发布

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最新推荐文章于 2022-07-28 16:41:37 发布

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分类专栏： Leetcode 文章标签： leetcode java

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本文介绍了一种使用二分查找算法来高效确定首次出现故障的产品版本的方法。在一个包含多个连续开发版本的产品中，需要找到最早引入故障的那个版本。通过定义一个API判断版本是否故障，并采用递归和非递归两种二分查找实现，可以显著减少检查次数。

【问题描述】

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.

【思路】

主要考察二分查找，二分查找有两种方式，一种的递归的，另外一种的非递归的循环实现。下面给出两种的代码。

注意：middle=(start+end)/2时会报出超时错误，因此需要写middle = start + (end - start) / 2

【code】

循环实现：

/* 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 start = 1, end = n;
 		while (start < end) {
 			int middle = start + (end-start) / 2;
			if (isBadVersion(middle)) {
 				end = middle;
 			}else {
 				start = middle+1;
			}
 		}
 		return start;
	}
}

递归实现：

private int searchRecursive(int start, int end) {
if (start == end) {
return start;
}
if (start < end) {
int middle = start + (end-start) / 2;
if (isBadVersion(middle)) {
return searchRecursive(start, middle);
} else {
return searchRecursive(middle+1, end);
}
}
return -1;
}