[LeetCode]127.Word Ladder

题目

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
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.

分析

广搜

代码

    /**------------------------------------    *   日期：2015-02-07    *   作者：SJF0115    *   题目: 127.Word Ladder    *   网址：https://oj.leetcode.com/problems/word-ladder/    *   结果：AC    *   来源：LeetCode    *   博客：    ---------------------------------------**/    #include <iostream>    #include <cstring>    #include <vector>    #include <queue>    #include <unordered_set>    using namespace std;    class Solution {    public:        int ladderLength(string start, string end, unordered_set<string> &dict) {            return BFS(start,end,dict);        }    private:        int BFS(string start,string end,unordered_set<string> &dict){            // 存放单词和单词所在层次            queue<pair<string,int> > q;            q.push(make_pair(start,1));            unordered_set<string> visited;            visited.insert(start);            // 广搜            bool found = false;            while(!q.empty()){                pair<string,int> cur = q.front();                q.pop();                string word = cur.first;                int len = word.size();                // 变换一位字符                for(int i = 0;i < len;++i){                    string newWord(word);                    for(int j = 0;j < 26;++j){                        newWord[i] = 'a' + j;                        if(newWord == end){                            found = true;                            return cur.second+1;                        }//if                        // 判断是否在字典中并且是否已经访问过                        if(dict.count(newWord) > 0 && visited.count(newWord) == 0){                            visited.insert(newWord);                            q.push(make_pair(newWord,cur.second+1));                        }//if                    }//for                }//for            }//while            if(!found){                return 0;            }//if        }    };    int main(){        Solution s;        unordered_set<string> set = {"hot","dot","dog","lot","log"};        //unordered_set<string> set = {"a","b","c"};        string start("hit");        string end("cog");        int result = s.ladderLength(start,end,set);        // 输出        cout<<result<<endl;        return 0;    }

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