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.

c++代码

每个数字逐一判断各个位置

class Solution {
    enum {
        S,V,I //Same: 0 1 8 Valid:2 5 6 9 Invalid:3 4 7
    };
public:
    int rotatedDigits(int N) {
        int i,num,count=0;
        int is_valid[]={S,S,V,I,I,V,V,I,S,V};
        bool found=false;
        for(int i=2;i<=N;i++){
            num=i;found=false;
            while(num){
                if(is_valid[num%10]==I){found=false;break;}
                if(is_valid[num%10]==V) found=true;
                num=num/10;
            }
            if(found==true) count++;
        }
        return count;     
    }
};

 动态规划

/*
dp[i] = 0, invalid number
dp[i] = 1, valid and same number
dp[i] = 2, valid and different number
*/
class Solution {
public:
    int rotatedDigits(int N) {
        vector<int> dp=vector<int>(N+1,0);
        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;
         
    }
};

 python代码

class Solution(object):
    def rotatedDigits(self, N):
        """
        :type N: int
        :rtype: int
        """
        dp=[0 for _ in xrange(N+1)]
        cnt=0
        for i in xrange(N+1):
            if i<10:
                if i==3 or i==4 or i==7:
                    dp[i]=0 #invalid
                if i==0 or i==1 or i==8:
                    dp[i]=1 #same
                if i==2 or i==5 or i==6 or i==9:
                    dp[i]=2 #valid
            else:
                dp[i]=min(dp[i%10]*dp[i/10],2)
            
            if dp[i]==2: 
                cnt+=1
        return cnt;