788. Rotated Digits

这里给出两种方法

方法一，逐个数字逐位判定，速度有点慢，但是很好理解

class Solution 
{
public:
    int rotatedDigits(int N) 
    {
        unordered_set<int> unvalid{3,4,7};
        unordered_set<int> change{2,5,6,9};
        int count=0;
        for(int i=2;i<=N;i++)
        {
            int cur=i;
            int flag=0;
            while(cur)
            {
                int last=cur%10;
                if(unvalid.find(last)!=unvalid.end())
                {
                    flag=0;
                    break;
                }
                if(change.find(last)!=change.end())
                    flag=1;
                cur/=10;
            }
            count+=flag;
        }
        return count;
    }
};

方法二，利用前面的判定结果，加快判定速度

class Solution 
{
public:
    int rotatedDigits(int N) 
    {
        vector<int> judge(N+1,0);
        int count=0;
        for(int i=0;i<=N;i++)
        {
            if(i<10)
            {
                if(i==0||i==1||i==8)
                    judge[i]=1;
                else if(i==3||i==4||i==7)
                    judge[i]=0;
                else
                {
                    judge[i]=2;
                    count++;
                }
            }
            else
            {
                int pre=judge[i/10];
                int last=judge[i%10];
                if(pre==1&&last==1)
                    judge[i]=1;
                else if(pre+last>2)
                {
                    judge[i]=2;
                    count++;
                }
                else
                    judge[i]=0;
            }
        }
        return count;
    }
};