(BFS 图的遍历) 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.

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'.

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这个题可能看起来很难，但是，如果理解了题意，就会发现这个题与迷宫遍历的问题，而且是最短路问题，四位数的加1和减1，可以看做迷宫的上下左右的方向，一共有8个“方向”，同时，题中还给了deadends，这个相当于迷宫中的障碍物，target就是迷宫中的出口，一次遍历，到最后没找到target，返回-1。

C++代码：

class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        unordered_set<string> dead(deadends.begin(),deadends.end());  //这个就是把deadends里面的字符串传给一个set集合。
        if(dead.count("0000")) return -1;  //0000可以看做迷宫的出发点，迷宫中的出发点是一个障碍，当然不能出发了。
        int res = 0;
        unordered_set<string> visit{{"0000"}};  //判断旋转的数字是否已经遍历了，相当于迷宫中的接下来的位置是否遍历了。
        queue<string> q{{"0000"}};
        while(!q.empty()){
            res++;
            for(int i = q.size(); i > 0; i--){  //要加这个循环。
                auto t = q.front();q.pop();
                for(int j = 0; j < t.size(); j++){
                    for(int k = -1; k <= 1; k++){
                        if(k == 0) continue;
                        string str = t;
                        str[j] = ((t[j] - '0') + 10 + k) % 10 + '0';  //为了让0-1最后得到9。
                        if(str == target) return res;
                        if(!visit.count(str) && !dead.count(str))
                            q.push(str);
                        visit.insert(str);
                    }
                }
            }
        }
        return -1;
    }
};