LWC 73: 788. Rotated Digits

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LWC 73: 788. Rotated Digits

传送门：788. Rotated Digits

Problem:

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：
观察N最大不超过10000，可以暴力解决。遍历1到N，核实每个num，如果num符合Rotated Digits则计数。

Java版本:

    public int rotatedDigits(int N) {
        int cnt = 0;
        for (int i = 1; i <= N; ++i) {
            int rotate = valid(i);
            if (rotate != -1 && rotate != i) cnt += 1;
        }
        return cnt;
    }

    public int valid(int n) {
        int[] map = {0, 1, 5, -1, -1, 2, 9, -1, 8, 6};
        int num = 0; int p = 1;
        for(;n != 0; n /= 10) {
            int digit = map[n % 10];
            if (digit == -1) return -1;
            num = num + p * digit;
            p *= 10;
        }
        return num;
    }

Python版本：

class Solution(object):
    def rotatedDigits(self, N):
        s1 = set([1, 8, 0])
        s2 = set([1, 2, 5, 8, 6, 9, 0])
        def isGood(n):
            s = set([int(i) for i in str(n)])
            return s.issubset(s2) and not s.issubset(s1)
        return sum(isGood(i) for i in range(N + 1))