難度 中等

題目 Leetcode:

752.Open the Lock

　　You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'. The wheels can rotate freely and wrap around: for example we can turn '9' to be '0', or '0' to be '9'. Each move consists of turning one wheel one slot.

　　The lock initially starts at '0000', a string representing the state of the 4 wheels.

　　You are given a list of deadends dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.

　　Given a target representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.

Constraints:

• 1 <=deadends.length <= 500
• deadends[i].length == 4
• target.length == 4
• target will not be in the list deadends.
• target and deadends[i] consist of digits only.

Keyword

slot狹槽

initially 最初

題目解析

本題大意就是要把0000轉到target，每次改變一位，但改變的過程中不能與死亡數字相同、一開始我是頭鐵用string陣列存每次情況，很快啊，直接爆了，然後就轉而去用int，我把次數存在每次情況裡

比如初始入隊10000代表第1次操作把0000放入佇列，出隊完入隊20001代表第二次操作入隊0001以此類推，但是這樣毫無疑問每次都需要取模運算量還是不小的，其實這題用二進位制存的話會很快。

 1 queue<int>x;
 2 bitset<100005>book;
 3 class Solution {
 4 public:
 5 int ans;
 6     bool judge(int temp,int nm[],int length)
 7     {
 8         int ttemp=temp%10000;
 9         for(int i=0;i<length;i++)
10         {
11             if(ttemp==nm[i])return 0;
12         }
13         return 1;
14     }
15     void record(int temp)
16     {
17         book[temp%10000]=1;
18         x.push(temp+10000);
19     }
20     bool bfs(int target,int nm[],int length)
21     {
22         while(!x.empty())
23         {
24             int a=x.front();
25             x.pop();
26             int temp;
27             if(a%10000==target)
28             {
29                 ans=a;
30                 return 1;
31             }
32             else
33             {
34                 for(int i=1;i<=4;i++)
35                 {
36                     temp=a;
37                     int cnt=pow(10,i);//10000
38                     int acnt=pow(10,i-1);
39                     if((temp % cnt)/acnt!=9) temp+=acnt;
40                     else temp-=9*acnt;
41                     if((!book[temp%10000])&&judge(temp,nm,length))record(temp);
42                     temp=a;
43                     if((temp % cnt)/acnt!=0) temp-=acnt;
44                     else temp+=9*acnt;
45                     if((!book[temp%10000])&&judge(temp,nm,length))record(temp);
46                 }
47             }
48         }
49         return 0;       
50     }
51     int openLock(vector<string>& deadends, string target) {
52         int nm[502];
53         stringstream ss;
54         ss<<target;
55         int ttarget;
56         ss>>ttarget;
57         int length=deadends.size();
58         for(int i=0;i<length;i++)
59         {
60             stringstream ss;
61             ss<<deadends[i];
62             ss>>nm[i];
63         }
64         while(!x.empty())x.pop();
65         x.push(10000);
66         book.reset();
67         book[0]=1;
68         if(!judge(10000,nm,length))return -1;
69         if(bfs(ttarget,nm,length))return ans/10000-1;
70         else return -1;
71     }
72 };