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

 1 class Solution 
 2 {
 3 public:
 4     int rotatedDigits(int N) 
 5     {
 6         unordered_set<int> unvalid{3,4,7};
 7         unordered_set<int> change{2,5,6,9};
 8         int count=0;
 9         for(int i=2;i<=N;i++)
10         {
11             int cur=i;
12             int flag=0;
13             while(cur)
14             {
15                 int last=cur%10;
16                 if(unvalid.find(last)!=unvalid.end())
17                 {
18                     flag=0;
19                     break;
20                 }
21                 if(change.find(last)!=change.end())
22                     flag=1;
23                 cur/=10;
24             }
25             count+=flag;
26         }
27         return count;
28     }
29 };

方法二，利用前面的判定结果，加快判定速度

 1 class Solution 
 2 {
 3 public:
 4     int rotatedDigits(int N) 
 5     {
 6         vector<int> judge(N+1,0);
 7         int count=0;
 8         for(int i=0;i<=N;i++)
 9         {
10             if(i<10)
11             {
12                 if(i==0||i==1||i==8)
13                     judge[i]=1;
14                 else if(i==3||i==4||i==7)
15                     judge[i]=0;
16                 else
17                 {
18                     judge[i]=2;
19                     count++;
20                 }
21             }
22             else
23             {
24                 int pre=judge[i/10];
25                 int last=judge[i%10];
26                 if(pre==1&&last==1)
27                     judge[i]=1;
28                 else if(pre+last>2)
29                 {
30                     judge[i]=2;
31                     count++;
32                 }
33                 else
34                     judge[i]=0;
35             }
36         }
37         return count;
38     }
39 };

 

转载于:https://www.cnblogs.com/zhuangbijingdeboke/p/9179214.html