LeetCode解题报告（392) -- 752. Open the Lock

Problem

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:

1 2	Input: deadends = ["0000"], target = "8888" Output: -1

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.

Analysis

题目的背景是开锁，从0000开始到题目给定的target，同时还给出了几个deadends状态。这个其实本质上就是一个图的问题，和二维矩阵中的地图是一个道理。deadends就是不能走的地方，转移的状态就是每个位置上+1和-1（0减1变9，9加1变0）。

使用BFS解决，用visited记录已经访问过的状态，每次生成新的状态好，首先要判断是否已经访问过，同时还需要判断是否是deadend，两者都为否才放入queue中。

Solution

这里处理状态转移有一个技巧，为了统一处理，可以先把数字加10，然后加上变化的+1或-1，然后再取个位数字。

Code

class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        string lock = "0000";
        unordered_set<string> deadlock(deadends.begin(), deadends.end());
        if (deadlock.count(lock)) {
            return -1;
        }
        unordered_set<string> visited({"0000"});
        if (visited.count(target)) {
            return 0;
        }
        queue<string> q;
        q.push(lock);
        int level = 0;
        
        while (!q.empty()) {
            level++;
            int size = q.size();
            for (int i = 0; i < size; i++) {
                string str = q.front();
                q.pop();
                
                for (int j = 0; j < str.size(); j++) {
                    for (int k = -1; k <= 1; k++) {
                        if (k == 0) {
                            continue;
                        }
                        string temp = str;
                        temp[j] = ((str[j] - '0') + 10 + k) % 10 + '0';
                        if (temp == target) {
                            return level;
                        }
                        if (!visited.count(temp) && !deadlock.count(temp)) {
                            q.push(temp);
                        }
                        visited.insert(temp);
                    }
                }
            }
        }
        return -1;
    }
};

Summary

这道题的本质还是BFS，只不过背景换成了字符串有点陌生。这道题目的分享到这里，感谢你的支持！