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
1classSolution {2public:3intladderLength(stringstart,stringend, unordered_set<string> &dict) {4if(dict.empty())return0;5if(start.empty() || end.empty())return0;6if(start.length() != end.length())return0;7dict.insert(end);89vector<unordered_set<string> > layers(2);10intcur =0, ...
q.push(word); } // word 是给定的 wordList 中不存在的单词,则跳过 } word[i] = ch; // 恢复 word[i]原来的值 } } } return 0; } }; 双向BFS c++实现std::swap(集合 1,集合 2);q.count(aChar) 计算包含元素的数量class Solution { public: int ladderLength(string beginWord, string end...
由于在找到一个转换路径后就返回,此返回值可以确保是最小值 1publicclassSolution {2publicstaticintladderLength(String start, String end,3HashSet<String>dict) {4intresult = 0;5if(dict.size() == 0) {6returnresult;7}89dict.add(start);10dict.add(end);1112result =BFS(start, end, dict);1314...
func ladderLength(beginWord string, endWord string, wordList []string) int { // 先把开始单词放入单词列表中,方便后续使用下标处理 wordList = append(wordList, beginWord) startIndex := len(wordList) - 1 // 找到结束单词在单词列表中的下标 endIndex := -1 for i, word := range wordList { ...
public int ladderLength(String beginWord, String endWord, Set<String> wordList) { Queue<Word> queue = new LinkedList<Word>(); queue.offer(new Word(beginWord, 1)); wordList.add(endWord); while (!queue.isEmpty()) { Word w = queue.poll(); ...
Word Ladder -- LeetCode 原题链接:http://oj.leetcode.com/problems/word-ladder/ 这道题看似一个关于字符串操作的题目,其实要解决这个问题得用图的方法。我们先给题目进行图的映射,顶点则是每个字符串,然后两个字符串如果相差一个字符则我们进行连边。接下来看看这个方法的优势,注意到我们的字符集只有小写字母...
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 Word Ladder Given two words (startandend), and a dictionary, find the length of shortest transformation sequence fromstarttoend, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary...
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