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].
将一个数字的每一位旋转180度，如果跟之前的数字不一样则是有效的
统计1到N中有多少这样的数字
3,4,7旋转后是无效的
C++（5ms）：
1 class Solution { 2 public: 3 int rotatedDigits(int N) { 4 int res = 0 ; 5 for(int i = 1 ; i <= N ; i++){ 6 if (isValid(i)){ 7 res++ ; 8 } 9 } 10 return res ; 11 } 12 13 bool isValid(int n){ 14 bool flag = false ; 15 while(n > 0){ 16 if (n % 10 == 2) flag = true ; 17 if (n % 10 == 5) flag = true ; 18 if (n % 10 == 6) flag = true ; 19 if (n % 10 == 9) flag = true ; 20 if (n % 10 == 3) return false ; 21 if (n % 10 == 4) return false ; 22 if (n % 10 == 7) return false ; 23 n /= 10 ; 24 } 25 return flag ; 26 } 27 };