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.
思路:找到数组中第一个坏的,直接二分即可。
1 // Forward declaration of isBadVersion API. 2 bool isBadVersion(int version); 3 4 class Solution { 5 public: 6 int firstBadVersion(int n) { 7 int l = 1, r = n; 8 while (l < r) { 9 int mid = ((r - l) >> 1) + l; //中间位置 10 if (isBadVersion(mid)) { //如果中间位置是坏的,那么要找的在前面(包括这个位置),所以r不能赋值为mid - 1 11 r = mid; 12 } else { //如果是好的,要找的必然在后面 13 l = mid + 1; 14 } 15 } 16 return l; //最后要找的数就是l指向的 17 } 18 };