BFS中的最小步数模型
因为只有四个数字，因此将四位字符串转换成数字，判断和变化都是对数字进行的

const int N = 10010;

class Solution {
public:

    bool stop[N];
    int dist[N];
    bool visited[N];
    int d[4];

    int turn(int s, int n, int dir) {
        if (!dir)   dir = -1; // +1 right, -1 left
        for (int i = 0; i < 4; i++) d[i] = s % 10, s = s / 10;
        d[n] = (d[n] + dir + 10) % 10;
        int ans = 0;
        for (int i = 3; i >= 0; i--)    ans = (ans * 10) + d[i];
        return ans;
    }

    int openLock(vector<string>& deadends, string target) {
        for (auto s : deadends) stop[stoi(s)] = true;
        int end = stoi(target);
        memset(dist, 0x3f, sizeof dist);
        memset(visited, 0, sizeof visited);

        queue<int> q;
        q.push(0);
        visited[0] = true;
        dist[0] = 0;
        while (q.size()) {
            auto t = q.front(); q.pop();
            if (stop[t])   continue;
            if (t == end) 
                return dist[t];

            for (int i = 0; i < 4; i++) 
                for (int j = 0; j < 2; j++) {
                    int ne = turn(t, i, j);
                    if (stop[ne])   continue;
                    if (visited[ne])    continue;
                    q.push(ne);
                    dist[ne] = dist[t] + 1;
                    visited[ne] = true;
                }
        }

        return -1;
    }
};