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].

找出<=给出数字的集合中，全部颠倒之后和原数字不相同的个数

思路很简单，就是如何判断一个数字颠倒之后是否和原来的不一样,for循环集合中所有的数字

    public int rotatedDigits(int N) {
        int result = 0;
        for(int i = 1; i <= N; i++){
            String istr = i + "";
            if(istr.indexOf("3") >= 0
              || istr.indexOf("4") >= 0
              || istr.indexOf("7") >= 0){
                continue;
            }else if(istr.indexOf("2") >= 0
                    || istr.indexOf("5") >= 0
                    || istr.indexOf("6") >= 0
                    || istr.indexOf("9") >= 0){
                result ++;
            }
        }
        return result;
    }