[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;
    }

运行时间

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