leetcode 278 Fisrt Bad Version
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 2018-03-03 22:15:10  

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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/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    // public int firstBadVersion(int n) {
    //     return helper(1, n);
    // }
    // public int helper(int start, int end){
    //     if(start == end)
    //         return start;
    //     int mid = (start + end) >> 1;
    //     if(isBadVersion(mid))
    //         return helper(start, mid);
    //     else
    //         return helper(mid+1, end);
    // }
    public int firstBadVersion(int n){
        int start = 1;
        int end = n;
        while(start < end){
            int mid = start + ((end - start)>>1);
            if(isBadVersion(mid))
                end = mid;
            else
                start = mid+1;
        }
        return start;
    }
}

  • 简单的二分查找

  • 要注意递归的话会超时

  • 同时，如果start end 都是超大的数的话，直接start+end可能会导致溢出，因此，用 start + （end - start）/2 。 同时，位操作相比于直接相除速度要快。因此

    mid = start + ((end - start)>>1)