73th LeetCode Weekly Contest Rotated Digits

X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X. A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other; 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number.

Now given a positive number N, how many numbers X from 1 to N are good?

Example:
Input: 10
Output: 4
Explanation:
There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.

Note:

• N  will be in range [1, 10000].

讲的是数字倒过来能不能变成其他数字啦，不是那种翻转哦。

那么1,8,0是不会变的，2,5,6,9会变，其他变不了，那么我们只需要判断存在2,5,6,9这些就好了

 class Solution {
 public:
     int flag(string str){
         int len=str.length();
         if(len==){
             for(int i=;i<len;i++){
                 if(str[i]==''||str[i]==''||str[i]==''||str[i]==''){
                     return ;
                 }else{
                     return ;
                 }
             }
         }
         int fit=,fis=;
         for(int i=;i<len;i++){
             if(str[i]==''||str[i]==''||str[i]==''||str[i]==''){
                   fit=;
             }else if(str[i]==''||str[i]==''||str[i]==''){
                 fis=;
             }else{
                 return ;
             }
         }
         if(fit==){
             return ;
         }else{
             return ;
         }
     }
     int rotatedDigits(int N) {
         int dis=;
         for(int i=;i<=N;i++){
             string Str="";
             int num=i;
             while(num){
                 Str+=((num%)+'');
                 num/=;
             }
             if(flag(Str)){
                   //  cout<<Str<<endl;
                 dis++;
             }
             //cout<<Str<<" "<<i<<endl;
         }
         return dis;
     }
 };