Question

剑指 Offer II 109. 开密码锁

题目描述

一个密码锁由 4 个环形拨轮组成，每个拨轮都有 10 个数字： '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' 。每个拨轮可以自由旋转：例如把 '9' 变为 '0'，'0' 变为 '9' 。每次旋转都只能旋转一个拨轮的一位数字。

锁的初始数字为 '0000' ，一个代表四个拨轮的数字的字符串。

列表 deadends 包含了一组死亡数字，一旦拨轮的数字和列表里的任何一个元素相同，这个锁将会被永久锁定，无法再被旋转。

字符串 target 代表可以解锁的数字，请给出解锁需要的最小旋转次数，如果无论如何不能解锁，返回 -1 。

 

示例 1:

输入：deadends = ["0201","0101","0102","1212","2002"], target = "0202"
输出：6
解释：
可能的移动序列为 "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202"。
注意 "0000" -> "0001" -> "0002" -> "0102" -> "0202" 这样的序列是不能解锁的，因为当拨动到 "0102" 时这个锁就会被锁定。

示例 2:

输入: deadends = ["8888"], target = "0009"
输出：1
解释：
把最后一位反向旋转一次即可 "0000" -> "0009"。

示例 3:

输入: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
输出：-1
解释：
无法旋转到目标数字且不被锁定。

示例 4:

输入: deadends = ["0000"], target = "8888"
输出：-1

 

提示：

• 1 <= deadends.length <= 500
• deadends[i].length == 4
• target.length == 4
• target 不在 deadends 之中
• target 和 deadends[i] 仅由若干位数字组成

 

注意：本题与主站 752 题相同： https://leetcode.cn/problems/open-the-lock/

解法

方法一

class Solution:
  def openLock(self, deadends: List[str], target: str) -> int:
    s = set(deadends)
    if target in s or '0000' in s:
      return -1
    if target == '0000':
      return 0

    def prev(c):
      return '9' if c == '0' else str(int(c) - 1)

    def next(c):
      return '0' if c == '9' else str(int(c) + 1)

    def get(t):
      res = []
      t = list(t)
      for i in range(4):
        c = t[i]
        t[i] = prev(c)
        res.append(''.join(t))
        t[i] = next(c)
        res.append(''.join(t))
        t[i] = c
      return res

    visited = set()
    q = deque([('0000', 0)])
    while q:
      status, step = q.popleft()
      for t in get(status):
        if t in visited or t in s:
          continue
        if t == target:
          return step + 1
        q.append((t, step + 1))
        visited.add(t)
    return -1

class Solution {
  public int openLock(String[] deadends, String target) {
    Set<String> s = new HashSet<>(Arrays.asList(deadends));
    if (s.contains(target) || s.contains("0000")) {
      return -1;
    }
    if (Objects.equals(target, "0000")) {
      return 0;
    }
    Set<String> visited = new HashSet<>();
    Deque<String> q = new ArrayDeque<>();
    q.offerLast("0000");
    int step = 0;
    while (!q.isEmpty()) {
      ++step;
      for (int i = 0, n = q.size(); i < n; ++i) {
        String status = q.pollFirst();
        for (String t : get(status)) {
          if (visited.contains(t) || s.contains(t)) {
            continue;
          }
          if (Objects.equals(t, target)) {
            return step;
          }
          q.offerLast(t);
          visited.add(t);
        }
      }
    }
    return -1;
  }

  private char prev(char c) {
    return c == '0' ? '9' : (char) (c - 1);
  }

  private char next(char c) {
    return c == '9' ? '0' : (char) (c + 1);
  }

  private List<String> get(String t) {
    List res = new ArrayList<>();
    char[] chars = t.toCharArray();
    for (int i = 0; i < 4; ++i) {
      char c = chars[i];
      chars[i] = prev(c);
      res.add(String.valueOf(chars));
      chars[i] = next(c);
      res.add(String.valueOf(chars));
      chars[i] = c;
    }
    return res;
  }
}

class Solution {
public:
  int openLock(vector<string>& deadends, string target) {
    unordered_set<string> s(deadends.begin(), deadends.end());
    if (s.count(target) || s.count("0000")) return -1;
    if (target == "0000") return 0;
    unordered_set<string> visited;
    queue<string> q;
    q.push("0000");
    int step = 0;
    while (!q.empty()) {
      ++step;
      for (int i = 0, n = q.size(); i < n; ++i) {
        string status = q.front();
        q.pop();
        for (auto t : get(status)) {
          if (visited.count(t) || s.count(t)) continue;
          if (t == target) return step;
          q.push(t);
          visited.insert(t);
        }
      }
    }
    return -1;
  }

  char prev(char c) {
    return c == '0' ? '9' : (char) (c - 1);
  }

  char next(char c) {
    return c == '9' ? '0' : (char) (c + 1);
  }

  vector<string> get(string& t) {
    vector<string> res;
    for (int i = 0; i < 4; ++i) {
      char c = t[i];
      t[i] = prev(c);
      res.push_back(t);
      t[i] = next(c);
      res.push_back(t);
      t[i] = c;
    }
    return res;
  }
};