https://leetcode.com/problems/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].

代码：

class Solution {

public:

    int rotatedDigits(int N) {

        int ans = 0;

        for(int i = 1; i <= N; i ++) {

            if(isgood(i)) ans ++;

        }

        return ans;

    }

    bool isgood(int x) {

        string s = to_string(x);

        bool flag = false;

        for(int i = 0; i < s.length(); i ++) {

            if(s[i] == '3' || s[i] == '7' || s[i] == '4') return false;

            if(s[i] == '2' || s[i] == '5' || s[i] == '6' || s[i] == '9') flag = true;

        }

        return flag;

    }

};

　　如果数字里出现 3 4 7 数字的话就不是好数字 逐位判断 还是 emmmm 阔以的吧