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…

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 in the word list. Note…

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 intermediate word must exist in the word list…

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 = "…

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 in the word list. Note…

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 in the word list.…

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…

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,直到最终到达另一端.本质上广度优先遍历图. 需要注意的是,拓展下一个单词时不能对字典进行枚举,因为字典里的单词太多.幸好单词本身都不长,所以直接枚举单词所有可能的变形,看看在dict中出现没有. 当然还不止这些,上面的做法仍然会超时,需要添加剪枝策略: 1. 如果某个单词在以前遍历过了,那么以后都不用再考…

题目如下: 解题思路:DFS或者BFS都行.本题的关键在于减少重复计算.我采用了两种方法:一是用字典dic_ladderlist记录每一个单词可以ladder的单词列表:另外是用dp数组记录从startword开始到wordlist每一个word的最小转换次数,这一点非常重要,可以过滤很多无效的运算. 代码如下: class Solution(object): def getLadderList(self, w,d): l = [] r = [] for i in xrange(26): l.a…