Formatted question description: https://leetcode.ca/all/788.html

# 788. Rotated Digits (Easy)

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].

Companies:

Related Topics:
String

## Solution 1.

// OJ: https://leetcode.com/problems/rotated-digits/

// Time: O(ND) where D is the count of digits in N
// Space: O(1)
class Solution {
private:
bool isGood(int N) {
bool good = false;
while (N) {
int d = N % 10;
if (d == 3 || d == 4 || d == 7) return false;
if (d == 2 || d == 5 || d == 6 || d == 9) good = true;
N /= 10;
}
return good;
}
public:
int rotatedDigits(int N) {
int cnt = 0;
for (int i = 1; i <= N; ++i) {
if (isGood(i)) ++cnt;
}
return cnt;
}
};


Java

class Solution {
public int rotatedDigits(int N) {
int count = 0;
for (int i = 1; i <= N; i++) {
String numStr = String.valueOf(i);
if (numStr.indexOf('3') >= 0 || numStr.indexOf('4') >= 0 || numStr.indexOf('7') >= 0)
continue;
if (numStr.indexOf('2') >= 0 || numStr.indexOf('5') >= 0 || numStr.indexOf('6') >= 0 || numStr.indexOf('9') >= 0)
count++;
}
return count;
}
}