Question

剑指 Offer II 108. 单词演变

题目描述

在字典（单词列表） wordList 中，从单词 beginWord_ _和 endWord 的 转换序列 是一个按下述规格形成的序列：

• 序列中第一个单词是 beginWord 。
• 序列中最后一个单词是 endWord 。
• 每次转换只能改变一个字母。
• 转换过程中的中间单词必须是字典 wordList 中的单词。

给定两个长度相同但内容不同的单词_ _beginWord_ _和 endWord 和一个字典 wordList ，找到从 beginWord 到 endWord 的 最短转换序列 中的 单词数目 。如果不存在这样的转换序列，返回 0。

 

示例 1：

输入：beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
输出：5
解释：一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。

示例 2：

输入：beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
输出：0
解释：endWord "cog" 不在字典中，所以无法进行转换。

 

提示：

• 1 <= beginWord.length <= 10
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• beginWord、endWord 和 wordList[i] 由小写英文字母组成
• beginWord != endWord
• wordList 中的所有字符串 互不相同

 

注意：本题与主站 127 题相同： https://leetcode.cn/problems/word-ladder/

解法

方法一

class Solution:
  def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
    words = set(wordList)
    q = deque([beginWord])
    ans = 1
    while q:
      n = len(q)
      for _ in range(n):
        s = q.popleft()
        s = list(s)
        for i in range(len(s)):
          ch = s[i]
          for j in range(26):
            s[i] = chr(ord('a') + j)
            t = ''.join(s)
            if t not in words:
              continue
            if t == endWord:
              return ans + 1
            q.append(t)
            words.remove(t)
          s[i] = ch
      ans += 1
    return 0

class Solution {

  public int ladderLength(String beginWord, String endWord, List<String> wordList) {
    Set<String> words = new HashSet<>(wordList);
    Queue<String> q = new LinkedList<>();
    q.offer(beginWord);
    int ans = 1;
    while (!q.isEmpty()) {
      for (int i = q.size(); i > 0; --i) {
        String s = q.poll();
        char[] chars = s.toCharArray();
        for (int j = 0; j < chars.length; ++j) {
          char ch = chars[j];
          for (char k = 'a'; k <= 'z'; ++k) {
            chars[j] = k;
            String t = new String(chars);
            if (!words.contains(t)) {
              continue;
            }
            if (endWord.equals(t)) {
              return ans + 1;
            }
            q.offer(t);
            words.remove(t);
          }
          chars[j] = ch;
        }
      }
      ++ans;
    }
    return 0;
  }
}

class Solution {
public:
  int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
    unordered_set<string> words(wordList.begin(), wordList.end());
    queue<string> q{{beginWord}};
    int ans = 1;
    while (!q.empty()) {
      for (int i = q.size(); i > 0; --i) {
        string s = q.front();
        q.pop();
        for (int j = 0; j < s.size(); ++j) {
          char ch = s[j];
          for (char k = 'a'; k <= 'z'; ++k) {
            s[j] = k;
            if (!words.count(s)) continue;
            if (s == endWord) return ans + 1;
            q.push(s);
            words.erase(s);
          }
          s[j] = ch;
        }
      }
      ++ans;
    }
    return 0;
  }
};

func ladderLength(beginWord string, endWord string, wordList []string) int {
  words := make(map[string]bool)
  for _, word := range wordList {
    words[word] = true
  }
  q := []string{beginWord}
  ans := 1
  for len(q) > 0 {
    for i := len(q); i > 0; i-- {
      s := q[0]
      q = q[1:]
      chars := []byte(s)
      for j := 0; j < len(chars); j++ {
        ch := chars[j]
        for k := 'a'; k <= 'z'; k++ {
          chars[j] = byte(k)
          t := string(chars)
          if !words[t] {
            continue
          }
          if t == endWord {
            return ans + 1
          }
          q = append(q, t)
          words[t] = false
        }
        chars[j] = ch
      }
    }
    ans++
  }
  return 0
}