【Leetcode】 Word Ladder

题目：

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

分析：

先把起点加到队列中, 然后每次将字典中与队首距离为1的字符串加进队列, 直到最后出队列的是终点字符串, 为确保终点字符串存在, 我们可以先在字典中加进去终点字符串。而在本题中，在寻找与一个字符串相距为1的的字典中另一个字符串时，如果一个个遍历字典消耗时间比较多, 每次时间复杂度是O(n).。在单个字符串不是很长的情况下, 一个个查看改变一个字符然后在字典中查看是否存在效率要更高些, 其时间复杂度是O(k log n), 其中k为单个字符串长度, n为字典长度。

Java版本：

 public int ladderLength(String beginWord, String endWord, Set<String> wordDict) {        Set<String> reached = new HashSet<String>();        reached.add(beginWord);        wordDict.add(endWord);        int distance = 1;        while (!reached.contains(endWord)) {            Set<String> toAdd = new HashSet<String>();            for (String each : reached) {                for (int i = 0; i < each.length(); i++) {                    char[] chars = each.toCharArray();                    for (char ch = 'a'; ch <= 'z'; ch++) {                        chars[i] = ch;                        String word = new String(chars);                        if (wordDict.contains(word)) {                            toAdd.add(word);                            wordDict.remove(word);                        }                    }                }            }            distance++;            if (toAdd.size() == 0) return 0;            reached = toAdd;        }        return distance;    }

C++版本：

class Solution {  public:      int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {          wordList.insert(endWord);          queue<pair<string, int>> que;          que.push(make_pair(beginWord, 1));          wordList.erase(wordList.find(beginWord));          while(!que.empty())          {              auto val = que.front();              que.pop();              if(val.first == endWord) return val.second;              for(int i =0; i< val.first.size(); i++)              {                  string str = val.first;                  for(int j = 0; j < 26; j++)                  {                      str[i] = 'a'+j;                      if(wordList.count(str) == 1)                      {                          que.push(make_pair(str, val.second+1));                          wordList.erase(str);                      }                  }              }          }          return 0;      }  };  

Python版本：

from collections import dequeclass Solution(object):    def ladderLength(self, beginWord, endWord, wordList):                def construct_dict(word_list):            d = {}            for word in word_list:                for i in range(len(word)):                    s = word[:i] + "_" + word[i+1:]                    d[s] = d.get(s, []) + [word]            return d                    def bfs_words(begin, end, dict_words):            queue, visited = deque([(begin, 1)]), set()            while queue:                word, steps = queue.popleft()                if word not in visited:                    visited.add(word)                    if word == end:                        return steps                    for i in range(len(word)):                        s = word[:i] + "_" + word[i+1:]                        neigh_words = dict_words.get(s, [])                        for neigh in neigh_words:                            if neigh not in visited:                                queue.append((neigh, steps + 1))            return 0                d = construct_dict(wordList | set([beginWord, endWord]))        return bfs_words(beginWord, endWord, d)