leetcode 752. Open the Lock

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本文介绍了一个典型的图搜索问题——开锁游戏。通过BFS算法解决如何从初始状态“0000”通过最少步骤达到目标状态的问题，同时避免遇到死胡同状态。文章详细解释了算法实现细节，并提供了一个C++代码示例。

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:

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

题目：初始字符串“0000”，每次只能变一位数字，而且是变成相邻的数字。问几次变换可以成目标字符串。其中有一些字符串是不能经过的。

1、BFS。

2、BFS需要hash表来存已经遍历过的点，而且是搜索快速的，所以用set。

3、重复代码写成函数。

class Solution {
public:
    int openLock(vector<string>& deadends, string target)
    {
        set<string> st;
        for (auto s : deadends)
            st.insert(s);
        queue<string> que;
        que.push("0000");
        int ret = 0;
        set<string> hash;
        hash.insert("0000");
        while (!que.empty())
        {
            int size = que.size(); //分层
            for (int i = 0; i < size; i++)
            {
                string top = que.front();
                que.pop();
                if (st.find(top) != st.end())   continue;
                if (top == target)  return ret;
                //变化插入
                for (int j = 0; j < 4; j ++)
                {
                    char c = top[j];
                    if (c == '0')
                    {
                        top[j] = '1';
                        if (hash.find(top) == hash.end())
                        MY_insert(top, que, hash);
                        top[j] = '9';
                        MY_insert(top, que, hash);
                    }
                    else if (c == '9')
                    {
                        top[j] = '0';
                        MY_insert(top, que, hash);
                        top[j] = '8';
                        MY_insert(top, que, hash);
                    }
                    else
                    {
                        top[j] = c + 1;
                        MY_insert(top, que, hash);
                        top[j] = c - 1;
                        MY_insert(top, que, hash);
                    }
                    top[j] = c;
                }
            }
            ret ++;
        }
        return -1;
    }
private:
    void MY_insert(string& top, queue<string>& que, set<string>& hash)
    {
        if (hash.find(top) == hash.end())
        {
            que.push(top);
            hash.insert(top);
        }
    }
};