#120 Word ladder

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:

1. Only one letter can be changed at a time
2. 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;    }};