---
# System prepended metadata

title: 752. Open the Lock

---

# 752. Open the Lock

## Description
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.

## Examples
### 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 cannot reach the target without getting stuck.

## 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.`

## 思路
讀完題目判斷下去這個解題本惰我來說太大了。上網找解答。[求救](https://blog.csdn.net/fuxuemingzhu/article/details/82703064)

他的解法使用 BFS 之中的 queue，先把 deadends 丟進 set 之中，很巧妙解決遇到 deadends 時的作法。

使用以下去檢查是否會進入 deadends：
```cpp
                if (visit.count(tmp))
                    continue;
```
以下使我思考很久的要從 0 開始或是 9 開始。
看完發現我對這提還有很多不熟悉，程式整體表現也不是很好，改天再來重寫。

```cpp
class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        queue<string> que;
        que.push("0000");
        int count = 0;
        set<string> visit;
        for (string& d : deadends) {
            visit.insert(d);
        }
        while (!que.empty()) {
            int size = que.size();
            for (int i = 0; i < size; i++) {
                string tmp = que.front();
                que.pop();
                if (visit.count(tmp))
                    continue;
                if (tmp == target)
                    return count;
                for (int j = 0; j < 4; j++) {
                    for (int k = -1; k < 2; k += 2) {
                        string next = tmp;
                        next[j] = '0' + (next[j] - '0' + k + 10) % 10;
                        que.push(next);
                    }
                }
                visit.insert(tmp);
            }
            count++;
        }
        return -1;
    }
};
```