zoukankan      html  css  js  c++  java
  • [LintCode] 796. 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

    Example 1:

    Given deadends = ["0201","0101","0102","1212","2002"], target = "0202"
    Return 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:

    Given deadends = ["8888"], target = "0009"
    Return 1
    
    Explanation:
    We can turn the last wheel in reverse to move from "0000" -> "0009".
    

    Notice

    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'.

     
    public class Solution {
        /**
         * @param deadends: the list of deadends
         * @param target: the value of the wheels that will unlock the lock
         * @return: the minimum total number of turns 
         */
        public int openLock(String[] deadends, String target) {
            String start = "0000";
            // Write your code here
            Set<String> deadSet = new HashSet<>();
            // edge case
            for (String cur : deadends) {
                if (cur.equals(start)) {
                    return -1;
                }
                deadSet.add(cur);
            }
            
            Queue<String> queue = new LinkedList<>();
            Set<String> visited = new HashSet<>();
            int step = 0;
            queue.offer(start);
            visited.add(start);
            while (!queue.isEmpty()) {
                int size = queue.size();
                while (size-- > 0) {
                    String cur = queue.poll();  
                    System.out.println(cur);
                    if (cur.equals(target)) {
                        return step;
                    }
                    for (String nxt: getNext(cur, visited, deadSet)) {
                        queue.offer(nxt);
                        visited.add(nxt);
                    }
                }
                step += 1;
            }
            return -1;
        }
        
        private List<String> getNext(String cur, Set<String> visited, Set<String> deadSet) {
            List<String> res = new ArrayList<>();
            for (int i = 0; i < cur.length(); i++) {
                String newStr = helper(cur, i, 1);
                if (!visited.contains(newStr) && !deadSet.contains(newStr)) {
                    res.add(newStr);
                }
                newStr = helper(cur, i, 9);
                if (!visited.contains(newStr) && !deadSet.contains(newStr)) {
                    res.add(newStr);
                }
            }
            return res;
        }
        
        private String helper(String cur, int i, int offset) {
            String newStr = cur.substring(0, i) + (char)((cur.charAt(i) - '0' + offset) % 10 + '0') + cur.substring(i + 1);
            return newStr;
        }
    }
  • 相关阅读:
    (二十九)动态单元格
    (二十八)QQ好友列表的展开收缩
    (二十七)QQ好友列表的实现
    (二十六)静态单元格(Cell)
    (二十五)键盘的设置与TextField细节处理
    poj 1734 Sightseeing trip
    BZOJ 2200: [Usaco2011 Jan]道路和航线
    LUOGU P1073 最优贸易
    poj 3662 Telephone Lines
    poj 3539 Elevator
  • 原文地址:https://www.cnblogs.com/xuanlu/p/12571950.html
Copyright © 2011-2022 走看看