https://leetcode.com/problems/word-ladder-ii/

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

  1. Only one letter can be changed at a time
  2. 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.
class Solution {
public:
void dfs(vector<vector<string> >& res, unordered_map<string, vector<string> >& fa, vector<string> load, string beginWord, string curWord) {
if(curWord == beginWord) {
reverse(load.begin(), load.end());
res.push_back(load);
reverse(load.begin(), load.end());
return;
} for(int i=; i<fa[curWord].size(); ++i) {
load.push_back(fa[curWord][i]);
dfs(res, fa, load, beginWord, fa[curWord][i]);
load.pop_back();
}
} vector<vector<string>> findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {
vector<vector<string> > res;
if(beginWord.compare(endWord) == ) return res; wordList.insert(endWord); unordered_map<string, vector<string> > fa;
unordered_set<string> vis;
unordered_set<string> lev;
unordered_set<string> next_lev; lev.insert(beginWord);
bool found = false; while(!lev.empty() && !found) {
if(lev.find(endWord) != lev.end()) found = true; for(auto str: lev) vis.insert(str); for(auto str : lev) {
for(int i=; i<str.length(); ++i) {
for(char ch = 'a'; ch <= 'z'; ++ch) {
if(str[i] != ch) {
string tmp = str;
tmp[i] = ch;
if(wordList.find(tmp) != wordList.end() && vis.find(tmp) == vis.end()) {
next_lev.insert(tmp);
fa[tmp].push_back(str);
}
}
}
}
} lev.clear();
swap(lev, next_lev);
} if(found) {
vector<string> load;
load.push_back(endWord);
dfs(res, fa, load, beginWord, endWord);
} return res;
}
};

leetcode@ [126] Word Ladder II (BFS + 层次遍历 + DFS)的更多相关文章

  1. [LeetCode] 126. Word Ladder II 词语阶梯 II

    Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformat ...

  2. LeetCode 126. Word Ladder II 单词接龙 II(C++/Java)

    题目: Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transfo ...

  3. Java for LeetCode 126 Word Ladder II 【HARD】

    Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from ...

  4. [LeetCode] 126. Word Ladder II 词语阶梯之二

    Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformat ...

  5. leetcode 126. Word Ladder II ----- java

    Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformat ...

  6. Leetcode#126 Word Ladder II

    原题地址 既然是求最短路径,可以考虑动归或广搜.这道题对字典直接进行动归是不现实的,因为字典里的单词非常多.只能选择广搜了. 思路也非常直观,从start或end开始,不断加入所有可到达的单词,直到最 ...

  7. leetcode 127. Word Ladder、126. Word Ladder II

    127. Word Ladder 这道题使用bfs来解决,每次将满足要求的变换单词加入队列中. wordSet用来记录当前词典中的单词,做一个单词变换生成一个新单词,都需要判断这个单词是否在词典中,不 ...

  8. 126. Word Ladder II(hard)

    126. Word Ladder II 题目 Given two words (beginWord and endWord), and a dictionary's word list, find a ...

  9. 【leetcode】Word Ladder II

      Word Ladder II Given two words (start and end), and a dictionary, find all shortest transformation ...

随机推荐

  1. spoj 346

    当一个数大于等于12  那分别处以2, 3, 4之后的和一定大于本身    但是直接递归会超时    然后发现有人用map存了   膜拜..... #include <cstdio> #i ...

  2. eclipse查看jar包中class的中文注释乱码问题的解决

    1,问题来源是在eclipse中直接查看springside的class(由eclipse自动反编译)里面注释的乱码问题: Preferences-General-Workspace-Text fil ...

  3. PHP获取APP客户端的IP地址的方法

    分析php获取客户端ip 用php能获取客户端ip,这个大家都知道,代码如下: /** * 获取客户端ip * @param number $type * @return string */ func ...

  4. zoj 3380 Patchouli's Spell Cards 概率DP

    题意:1-n个位置中,每个位置填一个数,问至少有l个数是相同的概率. 可以转化求最多有l-1个数是相同的. dp[i][j]表示前i个位置填充j个位置的方案数,并且要满足上面的条件. 则: dp[i] ...

  5. spring结合时,web.xml的配置

    <!-- 1. web.xml配置 <context-param> <param-name>webAppRootKey</param-name> <pa ...

  6. LINQ to PostgreSQL Tutorial

    原文 LINQ to PostgreSQL Tutorial This tutorial guides you through the process of creating a simple app ...

  7. SGU 168

    SGU 168,寻找矩阵中右上方,右方,下方最小的元素,采用动态规划解答. #include <iostream> #include <vector> #include < ...

  8. Android ListView避免多线程加载一个同一资源

    当我们的ListView中的Item包含图片,而且这些图片是同一资源,我们用多线程去加载图片,这时候可能就发生了这种情况. 比如线程是人,第一个人去做加载图片到缓存的工作,还没做好时第二个人要这同一张 ...

  9. 生成整数自增ID(集群主键生成服务)

        在集群的环境中,有这种场景     需要整数自增ID,这个整数要求一直自增,并且需要保证唯一性. Web服务器集群调用这个整数生成服务,然后根据各种规则,插入指定的数据库.        一般 ...

  10. python中的commands模块

    commands模块用于调用shell命令 有3中方法: commands.getstatus()   返回执行状态 commands.getoutput()   返回执行结果 commands.ge ...