• [LeetCode] Word Ladder


    Word Ladder

    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.

    解题思路:

    第一的想法是用一个图来记录相邻关系。然后将问题转化成图的最短路径问题。

    class Solution {
    public:
        int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
            if(beginWord == endWord){
                return 1;
            }
            if(isNeibor(beginWord, endWord)){
                return 2;
            }
            if(wordDict.find(beginWord)==wordDict.end()){
                wordDict.insert(beginWord);
            }
            if(wordDict.find(endWord)==wordDict.end()){
                wordDict.insert(endWord);
            }
            //将set转化成vector
            vector<string> wordVect;
            for(unordered_set<string>::iterator it = wordDict.begin(); it!=wordDict.end(); it++){
                wordVect.push_back(*it);
            }
            int len = wordVect.size();
            vector<vector<int>> graph(len, vector<int>());  //邻接表存储图
            int beginIndex = -1;
            int endIndex = -1;
            for(int i=0; i<len; i++){
                if(beginWord == wordVect[i]){
                    beginIndex = i;
                }
                if(endWord == wordVect[i]){
                    endIndex = i;
                }
                for(int j=i+1; j<len; j++){
                    if(isNeibor(wordVect[i], wordVect[j])){
                        graph[i].push_back(j);
                        graph[j].push_back(i);
                    }
                }
            }
            vector<int> pre(len, -1);       //某个节点的前驱节点
            pre[beginIndex] = beginIndex;
            queue<int> q({beginIndex});     //广度优先法
            bool flag = false;              //是否已经找到了路径
            while(!q.empty()){
                int c = q.front();
                q.pop();
                for(int i=0; i<graph[c].size(); i++){
                    if(pre[graph[c][i]]<0){
                        pre[graph[c][i]] = c;
                        q.push(graph[c][i]);
                        if(graph[c][i]==endIndex){
                            flag = true;
                            break;
                        }
                    }
                }
                if(flag){
                    break;
                }
            }
            if(!flag){
                return 0;
            }
            int result = 1;
            int c = endIndex;
            while(pre[c]!=c){
                result++;
                c = pre[c];
            }
            return result;
        }
        
        bool isNeibor(const string& str1, const string& str2){
            int len = str1.length();
            if(len!=str2.length()){
                return false;
            }
            int c = 0;
            for(int i=0; i<len; i++){
                if(str1[i]!=str2[i]){
                    c++;
                }
            }
            return c==1;
        }
    };

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  • 原文地址:https://www.cnblogs.com/llguanli/p/7064297.html
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