mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-25 17:50:26 +08:00
205 lines
24 KiB
JSON
205 lines
24 KiB
JSON
{
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"questionId": "1000030",
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"questionFrontendId": "面试题 17.22",
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"categoryTitle": "LCCI",
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"boundTopicId": 93504,
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"title": "Word Transformer LCCI",
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"content": "<p>Given two words of equal length that are in a dictionary, write a method to transform one word into another word by changing only one letter at a time. The new word you get in each step must be in the dictionary.</p>\r\n\r\n<p>Write code to return a possible transforming sequence. If there is more than one sequence, return any of them.</p>\r\n\r\n<p><strong>Example 1:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong>\r\nbeginWord = "hit",\r\nendWord = "cog",\r\nwordList = ["hot","dot","dog","lot","log","cog"]\r\n\r\n<strong>Output:</strong>\r\n["hit","hot","dot","lot","log","cog"]\r\n</pre>\r\n\r\n<p><strong>Example 2:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong>\r\nbeginWord = "hit"\r\nendWord = "cog"\r\nwordList = ["hot","dot","dog","lot","log"]\r\n\r\n<strong>Output: </strong>[]\r\n\r\n<strong>Explanation:</strong> <em>endWord</em> "cog" is not in the dictionary, so there's no possible transforming sequence.</pre>\r\n",
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"translatedTitle": "单词转换",
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"translatedContent": "<p>给定字典中的两个词，长度相等。写一个方法，把一个词转换成另一个词， 但是一次只能改变一个字符。每一步得到的新词都必须能在字典中找到。</p>\n\n<p>编写一个程序，返回一个可能的转换序列。如有多个可能的转换序列，你可以返回任何一个。</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><strong>输入:</strong>\nbeginWord = "hit",\nendWord = "cog",\nwordList = ["hot","dot","dog","lot","log","cog"]\n\n<strong>输出:</strong>\n["hit","hot","dot","lot","log","cog"]\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><strong>输入:</strong>\nbeginWord = "hit"\nendWord = "cog"\nwordList = ["hot","dot","dog","lot","log"]\n\n<strong>输出: </strong>[]\n\n<strong>解释:</strong> <em>endWord</em> "cog" 不在字典中，所以不存在符合要求的转换序列。</pre>\n",
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"name": "Breadth-First Search",
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"code": "class Solution {\n public List<String> findLadders(String beginWord, String endWord, List<String> wordList) {\n\n }\n}",
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"code": "class Solution(object):\n def findLadders(self, beginWord, endWord, wordList):\n \"\"\"\n :type beginWord: str\n :type endWord: str\n :type wordList: List[str]\n :rtype: List[str]\n \"\"\"",
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"code": "\n\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nchar** findLadders(char* beginWord, char* endWord, char** wordList, int wordListSize, int* returnSize){\n\n}\n",
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"code": "public class Solution {\n public IList<string> FindLadders(string beginWord, string endWord, IList<string> wordList) {\n\n }\n}",
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"code": "/**\n * @param {string} beginWord\n * @param {string} endWord\n * @param {string[]} wordList\n * @return {string[]}\n */\nvar findLadders = function(beginWord, endWord, wordList) {\n\n};",
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"code": "class Solution {\n\n /**\n * @param String $beginWord\n * @param String $endWord\n * @param String[] $wordList\n * @return String[]\n */\n function findLadders($beginWord, $endWord, $wordList) {\n\n }\n}",
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"code": "# @param {String} begin_word\n# @param {String} end_word\n# @param {String[]} word_list\n# @return {String[]}\ndef find_ladders(begin_word, end_word, word_list)\n\nend",
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"code": "impl Solution {\n pub fn find_ladders(begin_word: String, end_word: String, word_list: Vec<String>) -> Vec<String> {\n\n }\n}",
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"code": "-spec find_ladders(BeginWord :: unicode:unicode_binary(), EndWord :: unicode:unicode_binary(), WordList :: [unicode:unicode_binary()]) -> [unicode:unicode_binary()].\nfind_ladders(BeginWord, EndWord, WordList) ->\n .",
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"code": "defmodule Solution do\n @spec find_ladders(begin_word :: String.t, end_word :: String.t, word_list :: [String.t]) :: [String.t]\n def find_ladders(begin_word, end_word, word_list) do\n\n end\nend",
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"从一个蛮力的递归解法开始。只需要创建所有一次编辑的单词，检查它们是否在字典中，然后尝试该编辑路径。",
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"一旦你有了一个蛮力解法，就可以尝试找到一个更快的方法以得到所有一次编辑的有效单词。当绝大多数字符串都不是有效的字典单词时，你不会想创建所有一次编辑的字符串。",
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"为了快速得到编辑距离为1的有效单词，试着将字典中的单词以一种有效的方式进行分组。注意，b_ll形式的所有单词（如bill、ball、bell和bull）的编辑距离为1。然而，这些并不是仅有的编辑距离为1的单词。",
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"创建从通配符形式（如b_ll）到该通配符所匹配的所有单词的映射。然后，当你想要查找与bill相隔编辑距离为1的所有单词时，可以在映射中查找_ill、b_ll、bi_l和bil_。",
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"你之前的算法可能类似于深度优先搜索。你能使它更快吗?",
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"广度优先的搜索通常比深度优先的搜索要快。在最坏的情况下未必如此，但在很多情况下都是这样。为什么？你能找到更快的方法吗？",
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"如果同时从起始单词和目标单词开始进行广度优先搜索，结果会怎样？"
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],
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"sampleTestCase": "\"hit\"\n\"cog\"\n[\"hot\",\"dot\",\"dog\",\"lot\",\"log\",\"cog\"]",
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