leetcode 278 First Bad Version

最新推荐文章于 2024-11-06 09:58:18 发布

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本文介绍了一种算法，用于从一系列软件版本中找出首个出现故障的版本。通过使用二分搜索策略，该方法能够最小化调用质量检查API的次数。

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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

class Solution {
public:
    int firstBadVersion(int n) {
        int left = 1, right = n, mid = 0;
        
        while(left < right) {
            mid = left + (right-left)/2;
            
            if(isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        
        return left;
    }
};

int BinarySearch(int array[], int n, int value) {
	int left = 0;
	int right = n - 1;
	//如果上面这句是int right = n, 那么下面有两处需要修改,以保证一一对应
	//1.循环的条件是while(left < right)
	//2.循环内,当arr[middle] > value的时候, right = mid

	while(left <= right) { //循环条件，适时而变
		//求中间位置,使用移位的方法是为了防止溢出，同时也更高效
		int middle = left+((right-left)>>1);

		if(array[middle] > value) {
			right = middle-1; //right赋值，适时而变
		}else if(array[middle] < value) {
			left = middle+1;
		} else {
			//数组中不相等的情况更多
			return middle;
		}
	}
}