【Leetcode】127. Word Ladder（字符串阶梯）-蒲公英云

【Leetcode】127. Word Ladder（字符串阶梯）

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:

Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is nota transformed word.

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.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

题目大意：

给出开始和结束字符串，同时给出一个字符串字典，字典中所有的字符串都是不重复等长的。每次从上一个字符串只变动一个字节。求出最短次数，从beginWord到endWord。

解题思路：

DFS容易陷入无用的循环中，结果超时。应该使用BFS，将wordList转化为字典，没找到一个下一个符合的字符串，就要通过本位置的’a’-‘z’找出全部的符合条件的字符串加入queue。

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        // BFS
        unordered_set<string> dict(wordList.begin(), wordList.end());
        queue<string> todo;
        todo.push(beginWord);
        int ladder=1;
        while(!todo.empty()){
            int n = todo.size();
            for(int i=0;i<n;i++){
                string word = todo.front();
                todo.pop();
                if(word == endWord){
                    return ladder;
                }
                dict.erase(word);
                for(int j=0;j<word.size();j++){
                    char c=word[j];
                    for(int k=0;k<26;k++){
                        word[j] = 'a'+k;
                        if(dict.find(word)!=dict.end()){
                            todo.push(word);
                        }
                    }
                    word[j]=c;
                }
            }
            ladder++;
        }
        return 0;
    }
};