LeetCode 788. Rotated Digits

原题链接在这里：https://leetcode.com/problems/rotated-digits/

题目：

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].

题解：

Pay attention that the quesiton is asking to rotate each digit, it is not rotating the whoe number.

Thus, if digit is 3,4,7, then it is invalid.

If digit is 0, 1, 8, it is valid, also the same.

If digit is 2, 5, 6, 9, it is valid and different.

Check the number with dp, dp[i] represents if i is good, dp[i] = 0, invalid. dp[i] = 1, valid and same. dp[i] = 2, valid and different.

Check dp[i / 10] and dp[i % 10]. If both are 1, then dp[i] = 1. Else if both >= 1, which means there is one part = 2, then dp[i] = 2.

Time Complexity: O(N).

Space: O(N).

AC Java:  

class Solution {
    public int rotatedDigits(int N) {
        if(N < 0){
            return 0;
        }
        
        int res = 0;
        int [] dp = new int[N + 1];
        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;
                    res++;
                }           
            }else{
                int a = dp[i / 10];
                int b = dp[i % 10];
                if(a == 1 && b == 1){
                    dp[i] = 1;
                }else if(a >= 1 && b >= 1){
                    dp[i] = 2;
                    res++;
                }
            }
        }
        
        return res;
    }
}