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==1){
            for(int i=0;i<len;i++){
                if(str[i]=='2'||str[i]=='5'||str[i]=='6'||str[i]=='9'){
                    return 1;
                }else{
                    return 0;
                }
            }
        }
        int fit=0,fis=0;
        for(int i=0;i<len;i++){
            if(str[i]=='2'||str[i]=='5'||str[i]=='6'||str[i]=='9'){
                  fit=1;
            }else if(str[i]=='1'||str[i]=='0'||str[i]=='8'){
                fis=1;
            }else{
                return 0;
            }
        }
        if(fit==1){
            return 1;
        }else{
            return 0;
        }
    }
    int rotatedDigits(int N) {
        int dis=0;
        for(int i=1;i<=N;i++){
            string Str="";
            int num=i;
            while(num){
                Str+=((num%10)+'0');
                num/=10;
            }
            if(flag(Str)){
                  //  cout<<Str<<endl;
                dis++;
            }
            //cout<<Str<<" "<<i<<endl;
        }
        return dis;
    }
};