nexts[*it].push_back(word) : nexts[word].push_back(*it); } else if(!reach && dict.find(word) != dict.end()){ nextLevel.insert(word); isForward ? nexts[*it].push_back(word) : nexts[word].push_back(*it); } } *ch = tmp; } } return reach || findLaddersHelper(backward...
【leetcode】Word Ladder II Word Ladder II 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 For example, Given: start="hit" ...
if(!wordList.contains(endWord)){ return res; } List<String> item = new ArrayList<>(); item.add(beginWord); boolean[] visited = new boolean[wordList.size()]; //递归调用 dfs(beginWord,endWord,wordList,res,item,visited); //找出最短长度的解 int len = Integer.MAX_VALUE; for(int i =...
2 使用map<string, vector<string> > 保留父母节点,这样确保只有一个string子节点,而可以有多个父母节点vector<string> 形成path的时候,因为最终根节点肯定是start,所以一个孩子节点分别从多个父母节点回溯,都必然会到达start根节点 //2014-2-18 update vector<vector<string>> findLadders(string start, string end...
【leetcode】Word Ladder II Question : 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 [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...
用第一种方法,直接在word ladder 基础上修改,虽然能得到正确解,但在大数据下会超时,考虑优化。 很容易就发现,当word list集合过大的时候,越遍历到后面原来的做法仍然需要遍历整个wordlist,实际上是没有必要的。因为如果之后存在最优解,之前遍历过的结点不会出现在以后,否则会形成循环,而循环是可以删去的。 所以增...
[leetcode] wordladder ii problem: 代码:https://play.golang.org/p/qdmadQUcEC package main import ( "fmt" ) func main() { start := "hit" end := "cog" dict := []string{"hot", "dot", "dog", "lot", "log"} dict = append(dict, start, end)...
Can you solve this real interview question? Word Ladder - A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that: * Every adjacent pair of words diff
还有些数据范围不明确的题，还是以126.Word Ladder II为例，可以构造输入范围内的合法数据使标程OLE，...