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  • 0752. Open the Lock (M)

    Open the Lock (M)

    题目

    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 
    

    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.

    题意

    给一把4位密码锁,每1位可以值域位0-9,每次可以向前或向后转动1位,同时存在一些“死胡同”密码,如果转到了其中之一,则无法继续更改密码。问最少需要转动几次,能转到指定密码。

    思路

    一般求最短次数可以考虑抽象成图用BFS解决。这里每一个不同的密码可以抽象成图的一个结点,该结点的相邻结点为所有只改动一位能得到的密码,通过BFS求出最短路径即可。


    代码实现

    Java

    class Solution {
        public int openLock(String[] deadends, String target) {
            int step = 0;
            Queue<Integer> q = new LinkedList<>();
            Set<Integer> dead = new HashSet<>();
            Set<Integer> visited = new HashSet<>();
            int t = Integer.parseInt(target);
    
            for (String deadend : deadends) {
                dead.add(Integer.parseInt(deadend));
            }
    
            if (!dead.contains(0)) {
                q.offer(0);
                visited.add(0);
            }
            while (!q.isEmpty()) {
                int size = q.size();
                for (int i = 0; i < size; i++) {
                    int cur = q.poll();
                    if (cur == t) return step;
                    for (int index = 0; index < 4; index++) {
                        int next = plus(cur, index);
                        int prev = minus(cur, index);
                        if (!visited.contains(next) && !dead.contains(next)) {
                            q.offer(next);
                            visited.add(next);
                        }
                        if (!visited.contains(prev) && !dead.contains(prev)) {
                            q.offer(prev);
                            visited.add(prev);
                        }
                    }
                }
                step++;
            }
    
            return -1;
        }
    
        private int plus(int origin, int index) {
            int num = 0;
            int[] digits = new int[4];
            for (int i = 0; i < 4; i++) {
                digits[i] = origin % 10;
                origin /= 10;
            }
            digits[index] = digits[index] == 9 ? 0 : digits[index] + 1;
            for (int i = 3; i >= 0; i--) {
                num = num * 10 + digits[i];
            }
            return num;
        }
    
        private int minus(int origin, int index) {
            int num = 0;
            int[] digits = new int[4];
            for (int i = 0; i < 4; i++) {
                digits[i] = origin % 10;
                origin /= 10;
            }
            digits[index] = digits[index] == 0 ? 9 : digits[index] - 1;
            for (int i = 3; i >= 0; i--) {
                num = num * 10 + digits[i];
            }
            return num;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/14851572.html
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