word-ladder总结

---

title: word ladder总结

categories: LeetCode

tags:

• 算法
• LeetCode
  
  comments: true
  
  date: 2016-10-16 09:42:30

---

题意

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time

Each intermediate word must exist in the word list

For example,

Given:

beginWord = "hit"

endWord = "cog"

wordList = ["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.

思路

因为上一个词到下一个词之间，变化的只是一个字母，因此只需遍历所有字母，进行替换即可；

利用BFS的思想，将改变后的字符串存进队列中，进行下一次的循环。

题解

这个的题解有很多种，但是其实都是使用了最基本的BFS思想；

交换队列，计算层数

int ladderLength(string begin, string end, unordered_set<string>& wordList) {
    // 当前层，下一层
    queue<string> current, next;
    // 判重
    unordered_set<string> visited;
    // 层次
    int level = 0;
    bool found = false;

    // 有点像内联函数
    auto state_is_target = [&](const string &s) {
        return s == end;
    };
    auto state_extend = [&](const string &s) {
        vector<string> result;

        for (size_t i = 0; i < s.size(); ++i) {
            string new_word(s);
            for (char c = 'a'; c <= 'z'; c++) {
                if (c == new_word[i]) continue;
                swap(c, new_word[i]);
                if (wordList.count(new_word) > 0 && !visited.count(new_word)) {
                    result.push_back(new_word);
                    visited.insert(new_word);
                }
                swap(c, new_word[i]);
            }
        }

        return result;
    };

    current.push(begin);

    while (!current.empty() && !found) {
        ++level;
        while (!current.empty() && !found) {
            const string str = current.front();
            current.pop();

            const auto& new_states = state_extend(str);
            for (const auto& state : new_states) {
                next.push(state);
                if (state_is_target(state)) {
                    //找到
                    found = true;
                    break;
                }
            }
        }
        swap(next, current);
    }

    if (found) return level + 1;
    else return 0;
}

总结：

使用了匿名函数，其中命名的state_is_target,state_extend是函数模版，分别做的操作就是判断字符串是否相等，返回字符串BFS遍历的下一层的数组；

同时用了两层循环，外循环用于计算层数和交换队列，内循环用于计算找到结尾字符串；

一个队列

/**
 *  改变单词中的字母， 查看是否存在
 *
 *  @param word     <#word description#>
 *  @param wordDict <#wordDict description#>
 *  @param toVisit  <#toVisit description#>
 */
void addNextWords(string word, unordered_set<string>& wordDict, queue<string>& toVisit) {
    // 删除wordDict中的word，因为相同的没有用
    wordDict.erase(word);

    for (int i = 0; i < (int)word.length(); i++) {
        // 记录下word的字母
        char letter = word[i];
        // 变换word中每个字母，一个一个字母的查找是否匹配，因为只能替换一个字母
        for (int j = 0; j < 26; j++) {
            word[i] = 'a' + j;
            // 找到后加入访问过的队列，并从wordDict中删除
            if (wordDict.find(word) != wordDict.end()) {
                toVisit.push(word);
                wordDict.erase(word);
            }
        }
        // 变换回去
        word[i] = letter;
    }
}
/**
 *  使用BFS
 *
 *  @param beginWord <#beginWord description#>
 *  @param endWord   <#endWord description#>
 *  @param wordDict  <#wordDict description#>
 *
 *  @return <#return value description#>
 */
int ladderLength2(string beginWord, string endWord, unordered_set<string>& wordDict) {
    // 加载最后的单词
    wordDict.insert(endWord);
    queue<string> toVisit;
    addNextWords(beginWord, wordDict, toVisit);

    // 从2开始，是因为需要经历开始和结尾的两个字母
    int dist = 2;
    // toVisit中存储的是通过改变一个字母就可以到达wordDict中存在的字符串
    while (!toVisit.empty()) {
        int num = (int)toVisit.size();
        for (int i = 0; i < num; i++) {
            string word = toVisit.front();
            toVisit.pop();
            if (word == endWord) return dist;
            addNextWords(word, wordDict, toVisit);
        }
        dist++;
    }

    return 0;
}

总结：

同样是计算BFS的层数；

两个指针

/**
 *  使用两个指针和BFS
 *
 *  @param beginWord <#beginWord description#>
 *  @param endWord   <#endWord description#>
 *  @param wordDict  <#wordDict description#>
 *
 *  @return <#return value description#>
 */
int ladderLength3(string beginWord, string endWord, unordered_set<string>& wordDict) {
    unordered_set<string> head, tail, *phead, *ptail;
    head.insert(beginWord);
    tail.insert(endWord);

    int dist = 2;
    // 将phead指向更小size的unordered_set以便减少运行时间
    while (!head.empty() && !tail.empty()) {
        if (head.size() < tail.size()) {
            phead = &head;
            ptail = &tail;
        }
        else {
            phead = &tail;
            ptail = &head;
        }
        unordered_set<string> temp;
        for (auto iterator = phead->begin(); iterator != phead->end(); iterator++) {
            string word = *iterator;
            wordDict.erase(word);

            // 原理和上面一样，更换其中的字符，
            for (int i = 0; i < (int)word.length(); i++) {
                char letter = word[i];
                for (int k = 0; k < 26; k++) {
                    word[i] = 'a' + k;
                    if (ptail->find(word) != ptail->end())
                        return dist;
                    if (wordDict.find(word) != wordDict.end()) {
                        temp.insert(word);
                        wordDict.erase(word);
                    }
                }
                word[i] = letter;
            }
        }
        dist++;
        swap(*phead, temp);
    }
    return 0;
}

总结：

两个指针指向了unordered_set的head和tail，其做法和第一种方法是相通的，只不过一个用了unordered_set，同时在循环前面还加上了根据数组大小判断指针指向不同的数组；

使用类

class WordNode {
public:
    string word;
    int numSteps;

public:
    WordNode(string w, int n) {
        this->word = w;
        this->numSteps = n;
    }
};
/**
 *  使用类去处理记录数据,和前面的思想是类似的
 *
 *  @param beginWord <#beginWord description#>
 *  @param endWord   <#endWord description#>
 *  @param wordDict  <#wordDict description#>
 *
 *  @return <#return value description#>
 */
int ladderLength4(string beginWord, string endWord, unordered_set<string>& wordDict) {
    queue<WordNode*> queue;
    queue.push(new WordNode(beginWord, 1));

    wordDict.insert(endWord);

    while (!queue.empty()) {
        WordNode * top = queue.front();
        queue.pop();

        string word = top->word;
        if (word == endWord) {
            return top->numSteps;
        }

        for (int i = 0; i < word.length(); i++) {
            char temp = word[i];  //先记录后面还有改回来
            for (char c = 'a'; c <= 'z'; c++) {
                if (temp != c) {
                    word[i] = c;
                }

                if (wordDict.find(word) != wordDict.end()) {
                    queue.push(new WordNode(word, top->numSteps + 1));
                    wordDict.erase(word);
                }

            }
            word[i] = temp;
        }
    }
    return 0;
}

总结：

使用类去记录每个字符串的单前遍历的层数，本质还是计算层数，只不过用类去保存，这样的好处就是减少了一些计算代码；

题意

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:

Only one letter can be changed at a time

Each intermediate word must exist in the word list

For example,

Given:

beginWord = "hit"

endWord = "cog"

wordList = ["hot","dot","dog","lot","log"]

Return

[

["hit","hot","dot","dog","cog"],

["hit","hot","lot","log","cog"]

]

Note:

All words have the same length.

All words contain only lowercase alphabetic characters.

思路

肯定需要两个数组，一个用来存储当前已经存在的字符串，一个用来存放遍历结果的字符串，最后两者进行交换，直到遍历当前数组为空；还有一个重点在于可以利用邻接矩阵的思想，用map存储字符串和数组一一对应关系，最后根据顺序进行输出map即可；

代码

vector<string> temp_path;
vector<vector<string>> result_path;

/**
 *  生成路径（从后往前）
 *
 *  @param unordered_map<string <#unordered_map<string description#>
 *  @param path                 <#path description#>
 *  @param start                <#start description#>
 *  @param end                  <#end description#>
 */
void generatePath(unordered_map<string, unordered_set<string>>& path, const string& start, const string& end) {
    temp_path.push_back(start);

    if (start == end) {
        vector<string> ret = temp_path;
        reverse(ret.begin(), ret.end());
        result_path.push_back(ret);
        return ;
    }

    for (auto it = path[start].begin(); it != path[start].end(); it++) {
        generatePath(path, *it, end);
        temp_path.pop_back();
    }
}

/**
 *
 *
 *  @param beginWord <#beginWord description#>
 *  @param endWord   <#endWord description#>
 *  @param wordList  <#wordList description#>
 *
 *  @return <#return value description#>
 */
vector<vector<string>> findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {
    temp_path.clear();
    result_path.clear();

    unordered_set<string> current_step;
    unordered_set<string> next_step;

    unordered_map<string, unordered_set<string>> path;

    unordered_set<string> unvisited = wordList; // 记录还没有访问过的结点

    if (unvisited.count(beginWord) > 0)
        unvisited.erase(beginWord);

    // 未访问过的值增加结尾字符串，开始处加入起点字符串
    unvisited.insert(endWord);
    current_step.insert(beginWord);

    while (current_step.count(endWord) == 0 && unvisited.size() > 0) {
        // 遍历当前层的所有元素
        for (auto cur = current_step.begin(); cur != current_step.end(); cur++) {
            string word = *cur;
            // 给字符串中的每个字符都替换字母
            for (int i = 0; i < beginWord.size(); i++) {
                for (int j = 0; j < 26; j++) {
                    string tmp = word;
                    // 相等情况下
                    if (tmp[i] == 'a' + j) {
                        continue;
                    }

                    tmp[i] = 'a' + j;
                    // 表明字典中存在该字符串，可以到达下一步
                    if (unvisited.count(tmp) > 0) {
                        next_step.insert(tmp);
                        path[tmp].insert(word);
                    }
                }
            }
        }

        // 表明下一层中已经不存在任何没有访问过的
        if (next_step.empty()) {
            break;
        }

        // 清除所有没有访问过的值
        for (auto it = next_step.begin(); it != next_step.end(); it++) {
            unvisited.erase(*it);
        }

        current_step = next_step;
        next_step.clear();
    }

    // 存在结果
    if (current_step.count(endWord) > 0) {
        generatePath(path, endWord, beginWord);
    }

    return result_path;
}

/**
 *  建立树
 *
 *  @param forward              <#forward description#>
 *  @param backward             <#backward description#>
 *  @param dict                 <#dict description#>
 *  @param unordered_map<string <#unordered_map<string description#>
 *  @param tree                 <#tree description#>
 *  @param reversed             <#reversed description#>
 *
 *  @return <#return value description#>
 */
bool buildTree(unordered_set<string> &forward, unordered_set<string> &backward, unordered_set<string> &dict, unordered_map<string, vector<string> > &tree, bool reversed)
{
    if (forward.empty()) return false;
    // 如果开头的长度比结尾长，则从结尾开始找到开头
    if (forward.size() > backward.size())
        return buildTree(backward, forward, dict, tree, !reversed);

    // 存在开头和结尾则删除
    for (auto &word: forward) dict.erase(word);
    for (auto &word: backward) dict.erase(word);

    unordered_set<string> nextLevel; // 存储树的下一层
    bool done = false; //是否进一步搜索
    for (auto &it: forward) //从树的当前层进行遍历
    {
        string word = it;
        for (auto &c: word)
        {
            char c0 = c; //存储初始值
            for (c = 'a'; c <= 'z'; ++c)
            {
                if (c != c0)
                {
                    // 说明找到最终结果
                    if (backward.count(word))
                    {
                        done = true; //找到
                        !reversed ? tree[it].push_back(word) : tree[word].push_back(it); //keep the tree generation direction consistent;
                    }
                    // 说明还没找到最终结果但是在字典中找到
                    else if (!done && dict.count(word))
                    {
                        nextLevel.insert(word);
                        !reversed ? tree[it].push_back(word) : tree[word].push_back(it);
                    }
                }
            }
            c = c0; //restore the word;
        }
    }
    return done || buildTree(nextLevel, backward, dict, tree, reversed);
}
/**
 *  生成路径
 *
 *  @param beginWord            <#beginWord description#>
 *  @param endWord              <#endWord description#>
 *  @param unordered_map<string <#unordered_map<string description#>
 *  @param tree                 <#tree description#>
 *  @param path                 <#path description#>
 *  @param paths                <#paths description#>
 */
void getPath(string &beginWord, string &endWord, unordered_map<string, vector<string> > &tree, vector<string> &path, vector<vector<string> > &paths) //using reference can accelerate;
{
    if (beginWord == endWord) paths.push_back(path); //till the end;
    else
    {
        for (auto &it: tree[beginWord])
        {
            path.push_back(it);
            getPath(it, endWord, tree, path, paths); //DFS retrieving the path;
            path.pop_back();
        }
    }
}

vector<vector<string>> findLadders2(string beginWord, string endWord, unordered_set<string> &dict) {
    vector<vector<string> > paths;
    vector<string> path(1, beginWord);
    if (beginWord == endWord)
    {
        paths.push_back(path);
        return paths;
    }

    // 分别存储开头和结尾
    unordered_set<string> forward, backward;
    forward.insert(beginWord);
    backward.insert(endWord);

    unordered_map<string, vector<string> > tree;
    // 用来判断是从开头找到结尾还是从结尾找到开头，因为会从长度短的一方开始进行查找
    bool reversed = false;
    if (buildTree(forward, backward, dict, tree, reversed))
        getPath(beginWord, endWord, tree, path, paths);
    return paths;
}

bool test() {
    unordered_set<string> words = {"hot","dot","dog","lot","log"};
    string start = "hit", end = "cog";

    vector<vector<string>> result = findLadders2(start, end, words);

    for (auto i = 0; i < result.size(); ++i) {
        for (auto j = 0; j < result[0].size(); ++j) {
            cout << result[i][j] << "->";
        }
        cout << endl;
    }
    return true;
}

这两种方法其实是一样的；