leetcode—word ladder II 1.题目描述 Given two words (startandend),anda 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 ="hit" end...
找到father 路径后,从end开始往前dfs就可以得到所有的结果了 1classSolution {2public:3vector<vector<string>> findLadders(stringstart,stringend, unordered_set<string> &dict) {45vector<vector<string>>result;6unordered_set<string> unvisited=dict;78dict.insert(start);9dict.insert(end);1011unordered_map...
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 =...
【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 For example, Given: start="hit" end=...
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...
disMap.put(nextWord, disMap.get(word) + 1); queue.add(nextWord); } } } } return 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。
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
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] 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)...
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...