（Easy) Rotated Digits - LeetCode

Description:

X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone.

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 and become invalid.

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].

Accepted

32K

Submissions

58.1K

Solution:

class Solution {
    public int rotatedDigits(int N) {
        int count = 0;
        for(int i = 0; i<=N; i++  ){
            if(Check(i)){
                
                count++;
            }
        }
        
        return count; 
    }
    
    
    public boolean Check(int n){
        
        
        // 0-> 0
        // 1 -> 1
        // 2->5
        // 3 invalid
        // 4 invalid
        // 5 -> 2
        // 6-> 9
        // 7 invalid
        // 8 -> 8
        // 9 -> 6.
        int tmp = n;
        int digit = 0;
        int Sum = 0;
        int k = 0 ;
        
        do{
        
            digit = tmp%10;
         
            if(digit ==3||digit ==4||digit==7){
                return false;
            }
            
            switch(digit){
case 2:
    //;
    digit = 5;
    break;
case 5:
    //...;
    digit = 2;
    break;
   
case 6:
    //...;
    digit = 9;
    break;
case 9:
    //...;
    digit = 6;
    break;
} 
            Sum = Sum+ digit* (int )Math.pow(10,k);
            
            tmp = tmp/10;
                    
            k++;
        }
        while ( tmp>0);
    
    
        if (Sum != n){
            return true;
        }
        
        else{
            return false;
        }
    }
}