于是有了下面的这份代码: classSolution { public: vector<vector<string> > findLadders(stringstart,stringend, unordered_set<string> &dict) { // end typing your C/C++ solution below // DO NOT write int main() function //areslipan@163.com map<string,vector<string> > path; unordered_set<st...
3. 与Word Ladder1中的set不同,我们使用一个hasmap来记录路径中的单词,另外,每一个单词对应的value是一个string的list,记录到达此单词的所有的可能的路径。创建这些路径的方法是:把前节点的所有的路径加上当前单词,复制一份加过来即可。 REF:http://blog.csdn.net/whuwangyi/article/details/21611433 1publicclas...
把每一种可能性都走一遍,找到所有的解,然后求出解中最小长度的解,然后对解中的item进行删除操作,这样就能得到想要的结果。 class Solution { public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) { List<List<String>> res = new ArrayList<>(); if(!wordList...
双向搜索解决WordLadderII问题 一到源自于 LeetCode 上的题,链接 Word Ladder II - LeetCodeleetcode.com/problems/word-ladder-ii/description/ 根据题意,要找到 beginWord 到 endWord 的所有变换路径,双向搜索的思路是:依次填充以 beginWord 和 endWord 为终点的两个有向图,当两个有向图出现共同单词时,以共...
LeetCode: Word Ladder II [127] 【题目】 Given two words (startandend), and a dictionary, find all shortest transformation sequence(s) fromstarttoend, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary...
【Leetcode】Word Ladder II https://leetcode.com/problems/word-ladder-ii/ 题目: beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord toendWord, such that: Only one letter can be changed at a time...
(cur);if(end.equals(cur)){res.add(newArrayList<>(solution));}else{List<String>neighbours=nodeNeighbors.get(cur);for(String next:nodeNeighbors.get(cur)){if(distance.get(next)==distance.get(cur)+1){dfs(next,end,dict,nodeNeighbors,distance,solution,res);}}}solution.remove(solution.size()-...
ladder: String->Integer, 存储有从start到该string最短需要几步 map: String -> List<String> (adjacent graph), List<String>中的str修改一个字母可以到String publicclassSolution{Map<String,List<String>>map;List<List<String>>res;public List<List<String>>findLadders(String beginWord,String endWord,List...
class Solution { public: int ladderLength(string beginWord, string endWord, vector<string>& wordList) { unordered_set<string> dict(wordList.begin(), wordList.end()); if (!dict.count(endWord)) return 0; queue<string> q; q.push(beginWord); int l = beginWord.length(); int step = 0...
Word Ladder II 一开始以为做了I之后,用一样的思路解决II应该会容易很多。但做了才发现,II对效率的要求更高,而且这使得这一题成为了LeetCode上通过率最低的题之一。 先讲一下自己做的时候的思路,和I相似,用图的思想,只不过是该用DFS(深度优先搜索),从起点出发,对字典中的每一个可达字符串进行DFS并记录路径...