【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

First Bad Version

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.

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C++   // Forward declaration of isBadVersion API. bool isBadVersion ( int version ); //序列可看成（0,0,0,0,1,1,1,1,1,1,1,1,1,1）找第一個為1的數（可用lower_bound函式） class Solution { public :     int firstBadVersion ( int n ) //版本序列為有序序列，可用二分查詢     {         int left = 1 , mid , right = n ;         while ( left <= right ) //這裡也可以寫成<         {             mid = left + ( right - left )/ 2 ; //防止（left+right）/2造成溢位風險             if (! isBadVersion ( mid ))             {                 left = mid + 1 ; //為0時往右邊找，left會移動到第一個為1的數的位置             }             else             {                 right = mid - 1 ; //為1時往左邊找，也可以寫成right=mid             }         }                 if ( isBadVersion ( left )) return left ;         else return - 1 ; //如果找不到就返回-1             } }; /*     //一般的二分查詢         while(left <= right) //注意這裡為<=，因為要計算一次mid再返回         {             mid = left + (right - left)/2; //防止（left+right）/2造成溢位風險             if(a[mid]<key)             {                 left = mid + 1;             }             else if(a[mid]>key)             {                 right = mid -1;             }             else             {                 return mid;             }         }         */