【Leetcode】127. Word Ladder（字符串阶梯）

╰+哭是因爲堅強的太久メ 2022-02-04 13:27 140阅读 0赞

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.

**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;
        }
    };