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]
.
Solution:
Brute-force. Check each number from 1 to N and see if it's a good number.
Code (Java):
class Solution { public int rotatedDigits(int N) { int count = 0; for (int num = 1; num <= N; num++) { if (isGoodNumber(num)) { count++; } } return count; } private boolean isGoodNumber(int num) { boolean has2569 = false; while (num > 0) { int digit = num % 10; has2569 |= contains2569(digit); if (contains347(digit)) { return false; } num /= 10; } return has2569; } private boolean contains347(int digit) { if (digit == 3 || digit == 4 || digit == 7) { return true; } return false; } private boolean contains2569(int digit) { if (digit == 2 || digit == 5 || digit == 6 || digit == 9) { return true; } return false; } }
No comments:
Post a Comment