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.  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.

fuck！没有看懂题目！！！听说是 一个数 valid 的条件是这个数不包含 3, 4, 7 且包含至少一个 2, 5, 6, 9 

最直观的解法：

class Solution(object):
    def rotatedDigits(self, N):
        """
        :type N: int
        :rtype: int
        """
        # 一个数 valid 的条件是这个数不包含 3, 4, 7 且包含至少一个 2, 5, 6, 9                        
        def is_valid(n):        
            found = False
            while n:
                d = n % 10
                if d in {3,4,7}:
                    return False
                if d in {2,5,6,9}:
                    found = True
                n /= 10
            return found
        
        ans = 0
        for n in range(1, N+1):
            if is_valid(n):
                ans += 1
        return ans

看还有dp解法的：

Using a int[] for dp.
dp[i] = 0, invalid number
dp[i] = 1, valid and same number
dp[i] = 2, valid and different number

    public int rotatedDigits(int N) {
        int[] dp = new int[N + 1];
        int count = 0;
        for(int i = 0; i <= N; i++){
            if(i < 10){
                if(i == 0 || i == 1 || i == 8) dp[i] = 1;
                else if(i == 2 || i == 5 || i == 6 || i == 9){
                    dp[i] = 2;
                    count++;
                }
            } else {
                int a = dp[i / 10], b = dp[i % 10];
                if(a == 1 && b == 1) dp[i] = 1;
                else if(a >= 1 && b >= 1){
                    dp[i] = 2;
                    count++;
                }
            }
        }
        return count;
    }

 最直观的解法：

class Solution(object):
    def rotatedDigits(self, N):
        counts = 0
        for num in range(1, N+1):
            number = str(num)
            if '3' in number or '7' in number or '4' in number: # This will be an invalid number upon rotation
                continue # Skip this number and go to next iteration
            if '2' in number or '5' in number or '6' in number or '9' in number:
                counts += 1
        return counts