思路：
相当于转化成查找第一个true，直接二分搜索即可。
简单题我重拳出击
代码：

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

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