[LeetCode] Word Ladder 词语阶梯

 

Given two words (beginWord and endWord), and a dictionary, 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 intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

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.

 

这道词句阶梯的问题给了我们一个单词字典，里面有一系列很相似的单词，然后给了一个起始单词和一个结束单词，每次变换只能改变一个单词，并且中间过程的单词都必须是单词字典中的单词，让我们求出最短的变化序列的长度。这道题还是挺有难度的，我当然是看了别人的解法才写出来的，这里用到了两个高级数据结构unordered_map和queue，即哈希表和队列，其中哈希表是记录单词和目前序列长度之间的映射，而队列的作用是保存每一个要展开的单词。首先把起始单词映射为1，把起始单词排入队列中，开始队列的循环，取出队首词，然后对其每个位置上的字符，用26个字母进行替换，如果此时和结尾单词相同了，就可以返回取出词在哈希表中的值加一。如果替换词在字典中存在但在哈希表中不存在，则将替换词排入队列中，并在哈希表中的值映射为之前取出词加一。如果循环完成则返回0，参见代码如下：

 

class Solution {
public:
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
        unordered_map<string, int> m;
        queue<string> q;
        m[beginWord] = 1;
        q.push(beginWord);
        while (!q.empty()) {
            string word = q.front(); q.pop();
            for (int i = 0; i < word.size(); ++i) {
                string newWord = word;
                for (char ch = 'a'; ch <= 'z'; ++ch) {
                    newWord[i] = ch;
                    if (newWord == endWord) return m[word] + 1;
                    if (wordDict.find(newWord) != wordDict.end() && m.find(newWord) == m.end()) {
                        q.push(newWord);
                        m[newWord] = m[word] + 1;
                    }   
                }
            }
        }
        return 0;
    }
};

 

参考资料：

https://leetcode.com/discuss/11648/a-25-line-1204ms-bfs-solution

http://www.cnblogs.com/TenosDoIt/p/3443512.html

 

LeetCode All in One 题目讲解汇总(持续更新中...)