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‘

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

Example 1:

Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202"
Output: 6
Explanation:
A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202".
Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid,
because the wheels of the lock become stuck after the display becomes the dead end "0102".

Example 2:

Input: deadends = ["8888"], target = "0009"
Output: 1
Explanation:
We can turn the last wheel in reverse to move from "0000" -> "0009".

Example 3:

Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
Output: -1
Explanation:
We can‘t reach the target without getting stuck.

Example 4:

Input: deadends = ["0000"], target = "8888"
Output: -1

Note:

1. The length of deadends will be in the range [1, 500].
2. target will not be in the list deadends.
3. Every string in deadends and the string target will be a string of 4 digits from the 10,000 possibilities ‘0000‘ to ‘9999‘.

關鍵詞"...return the minimum total number of turns...", 需要用BFS，用兩個queue，queue1當前層，queue2下一層。為避免重復，需要用哈希表存visited。

 1 class Solution {
 2 public:
 3     int openLock(vector<string>& deadends, string target) {
 4         unordered_set<string>deads(deadends.begin(),deadends.end());
 5         if(deads.find("0000")!=deads.end()){
 6             return -1;
 7         }
 8         int length = 0;
 9         queue<string> queue1;
10         queue<string> queue2;
11         unordered_set<string> visited;
12         queue1.push("0000");
13         visited.insert("0000");
14         while(!queue1.empty()){
15             string curr = queue1.front();
16             queue1.pop();
17             if(curr==target){
18                 return length;
19             }
20             for(int i=0;i<4;i++){
21                 char old = curr[i];
22                 //next code by changing on ith digit
23                 curr[i] = (old-‘0‘+1)%10+‘0‘;
24                 if(visited.find(curr)==visited.end()&&deads.find(curr)==deads.end()){ // new code
25                     queue2.push(curr);
26                     visited.insert(curr);
27                 }
28                 
29                 //previous code by changing on ith digit
30                 curr[i] = (old-‘0‘+9)%10+‘0‘;
31                 if(visited.find(curr)==visited.end()&&deads.find(curr)==deads.end()){ // new code
32                     queue2.push(curr);
33                     visited.insert(curr);
34                 }
35                 
36                 //restore current code
37                 curr[i]=old;
38             }
39                 
40             if(queue1.empty()){
41                 queue1=queue2;
42                 queue<string> tmp;
43                 queue2=tmp;
44                 length++;
45             }
46         }
47         
48         return -1;
49     }
50 };

752. Open the Lock