zoukankan      html  css  js  c++  java
  • [LC] 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".

    class Solution {
        public int openLock(String[] deadends, String target) {
            Set<String> deadSet = new HashSet<>();
            for (String str: deadends) {
                // start string might is deadend
                if ("0000".equals(str)) {
                    return -1;
                }
                deadSet.add(str);
            }
            Queue<String> queue = new LinkedList<>();
            Set<String> visited = new HashSet<>();
            int res = 0;
            queue.offer("0000");
            visited.add("0000");
            while (!queue.isEmpty()) {
                int size = queue.size();
                while (size-- > 0) {
                    String cur = queue.poll();
                    if (cur.equals(target)) {
                        return res;
                    }
                    for (int i = 0; i < 4; i++) {
                        String next = cur.substring(0, i) + (char)((cur.charAt(i) - '0' + 1) % 10  + '0') + cur.substring(i + 1);
                        if (!visited.contains(next) && !deadSet.contains(next)) {
                            visited.add(next);
                            queue.add(next);
                        }
                        next = cur.substring(0, i) + (char)((cur.charAt(i) - '0' + 9) % 10  + '0') + cur.substring(i + 1);
                        if (!visited.contains(next) && !deadSet.contains(next)) {
                            visited.add(next);
                            queue.add(next);
                        }
                    }
                }
                res += 1;
            }
            return -1;
        }
    }
  • 相关阅读:
    Java 并发编程(一):简介
    @程序员,你在颓丧的时候怎么办?
    @程序员,你该如何磨快你的锯子
    Java Socket:飞鸽传书的网络套接字
    Java -- JDBC 学习--使用 DBUtils
    Java -- JDBC 学习--数据库连接池
    Java -- JDBC 学习--批量处理
    Java -- JDBC 学习--事务
    Java -- JDBC 学习--处理Blob
    Java -- JDBC_DAO 设计模式
  • 原文地址:https://www.cnblogs.com/xuanlu/p/13059856.html
Copyright © 2011-2022 走看看