Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) 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 = "hi…

  Word Ladder II Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) 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, G…

Word Ladder 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,…

Word Ladder II Total Accepted: 11755 Total Submissions: 102776My Submissions Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each i…

[leetcode]Word Ladder II @ Python 原题地址:http://oj.leetcode.com/problems/word-ladder-ii/ 参考文献:http://blog.csdn.net/doc_sgl/article/details/13341405   http://chaoren.is-programmer.com/ 题意:给定start单词,end单词,以及一个dict字典.要求找出start到end的所有最短路径,路径上的每个单词都要出现在dict…

Word Ladder II Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a timeEach intermediate word must exist in the dictionaryFor example, Given…

Word Ladder II 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/word-ladder-ii/description/ Description Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such…

126. Word Ladder II 题目 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 transformed word must exist…

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,不在词典中就不能加入队列. pathCnt用来记录遍历到的某一个词使用的次数,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pathCnt中,如果在,则说明之前已经达到过,这次不用再计算了,因为这次计算的path肯定比之前多.pathCnt相当于剪枝. class Solution { pub…

增加了可以在构造Hangman对象时通过传入参数设定“最大猜测次数”的功能.少量修改.# 2015年12月15日 00:20:22 https://github.com/shalliestera/hangman // 猜单词游戏 #ifndef HANGMAN_H_ #define HANGMAN_H_ #include <string> class Hangman { public: // 可以在创建时确定最大猜错次数 explicit Hangman(int chances = 6); /…