#120 Word ladder

最新推荐文章于 2024-09-08 21:02:59 发布

原创最新推荐文章于 2024-09-08 21:02:59 发布 · 282 阅读

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#lintcode #BFS #word ladder

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本文介绍了一种使用广度优先搜索(BFS)算法解决单词从起始状态转换到目标状态的最短路径问题的方法。通过具体示例展示了如何在给定字典中找到最短转换路径。

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题目描述：

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the dictionary

Notice

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

Have you met this question in a real interview?

Yes

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.

题目思路：

寻求path的问题，在我看来dfs和bfs都可以。此题为寻找最短路径，有两种方法：dfs求出所有符合条件的路径，然后取最小值；bfs直接求得的是最短路径之一。此题用dfs会在dict较大时超时，而题目只是要求最短路径，所以bfs为最佳答案。

Mycode （AC = 252ms）：

class Solution {
public:
    /**
      * @param start, a string
      * @param end, a string
      * @param dict, a set of string
      * @return an integer
      */
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        // write your code here
        // consider the special case
        if (start == end) return 1;
        
        // add start/end into dict, in case dict doesn't contain them
        dict.insert(start);
        dict.insert(end);
        
        
        // bfs to search the smallest length
        queue<string> str_queue;
        queue<int> levels;
        
        str_queue.push(start);
        levels.push(1);
        while (!str_queue.empty()) {
            string crt = str_queue.front();
            str_queue.pop();
            
            int level = levels.front();
            levels.pop();
            
            vector<string> next_words = getNextWord(crt, dict);
            for (int i = 0; i < next_words.size(); i++) {
                string word = next_words[i];
                
                if (word == end) {
                    return level + 1;
                }
                else {
                    dict.erase(dict.find(word));
                    str_queue.push(word);
                    levels.push(level + 1);
                }
            }
            
        }
        
        return 0;
        
    }
    
    
    // get the next word that in the dict
    vector<string> getNextWord(string crt, unordered_set<string> &dict) {
        vector<string> next_word;
        
        for (int i = 0; i < crt.length(); i++) {
            for (char ch = 'a'; ch <= 'z'; ch++) {
                string tmp = crt;
                tmp[i] = ch;
                
                if (tmp != crt && dict.find(tmp) != dict.end()) {
                    next_word.push_back(tmp);
                }
            }
        }
        
        return next_word;
    }
};