[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. 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].

思路：

将区间内每一个数都依次判断是否是good number。

对于一个数，比如125，变换后为152，则是good number。判断每一位的情况。

map<int,int>mp;

bool isgood(int n)
{
    int temp = n;
    int t = 0,newnum = 0;    
    //if  vailid
    int power = 0;
    while(n > 0)
    {
        t = n % 10;
        n /= 10;        
        if( t == 3 || t == 4 || t == 7)return false;
        t = mp[t];
        newnum += t * pow(10,power);
        power++;
    }    
    if(temp != newnum)return true;
    return false;
}
int rotatedDigits(int N)
{mp[0] = 0,mp[1] = 1,mp[8] = 8,mp[2] = 5,mp[5] = 2,mp[6] = 9,mp[9] = 6;
    int cnt = 0;
    for(int i = 1;i <= N;i++)
    if(isgood(i))cnt++;
    return cnt;
}