LeetCode|788. 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. 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].

Solution

本题题意很容易理解，0到9的数字如果这个数字旋转180°之后还是数字那么这个数字叫做旋转数，如果旋转180°后还是自身那么这个数可以称为自旋数，如果旋转后不是数字那么这个数就是非旋转数。这里面有个点题意没说，也就是说如果多位数字N只要包含一个非旋转数，则N为非旋转数。换句话说，旋转数中只能包含自旋数和旋转数。

我们用代码很容易实现这个思路。

class Solution {
    static int[] index={0,0,1,-1,-1,1,1,-1,0,9};
    public int rotatedDigits(int N) {
        int count=0;
        for(int i=0;i<=N;i++){
            if(isRotate(i)){
                count++;
            }
        }
        return count;
    }
    
        private static boolean isRotate(int num){
        boolean result=false;
        int selfCount=0;
        int count=0;
        while(num>0){
            int cur=num%10;
            count++;
      if(index[cur]==-1){
                result=false;
                break;
            }else if(index[cur]==0){
                selfCount++;
            }else{
                result=true;
            }
            num=num/10;
        }
        return result&&selfCount!=count;
    }
}