diff --git a/Week_04/id_9/LeetCode_211_9.cs b/Week_04/id_9/LeetCode_211_9.cs new file mode 100644 index 00000000..4d97bad3 --- /dev/null +++ b/Week_04/id_9/LeetCode_211_9.cs @@ -0,0 +1,103 @@ +using System; +using System.Collections.Generic; +using System.Linq; +using System.Text; +using System.Threading.Tasks; + +namespace ProblemSolutions +{ + public class Problem211 : IProblem + { + public void RunProblem() + { + WordDictionary obj = new WordDictionary(); + obj.AddWord("bad"); + obj.AddWord("dad"); + obj.AddWord("mad"); + bool param_2 = obj.Search("pad"); + param_2 = obj.Search("bad"); + param_2 = obj.Search(".ad"); + param_2 = obj.Search("b.."); + param_2 = obj.Search("b.d"); + } + + public class WordDictionary + { + public class TrieNode + { + public TrieNode(char c) + { + m_char = c; + } + public char m_char; + public bool m_isWord; + public TrieNode[] m_nextNode = new TrieNode[26]; + } + + private TrieNode rootNode = new TrieNode('\\'); + + /** Initialize your data structure here. */ + public WordDictionary() + { + + } + + /** Adds a word into the data structure. */ + public void AddWord(string word) + { + var nodeTemp = rootNode; + foreach (var wordItem in word) + { + var charIndex = wordItem - 'a'; + if (nodeTemp.m_nextNode[charIndex] == null) + nodeTemp.m_nextNode[charIndex] = new TrieNode(wordItem); + + nodeTemp = nodeTemp.m_nextNode[charIndex]; + } + nodeTemp.m_isWord = true; + } + + /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */ + public bool Search(string word) + { + return RecurSive(rootNode, word, -1); + } + + private bool RecurSive(TrieNode node, string word, int charIndex) + { + //查看当前节点满足的条件 + if (charIndex > word.Length - 1) return false; + if (charIndex == word.Length - 1) return node.m_isWord && (node.m_char == word[charIndex] || word[charIndex] == '.'); + + //查看下一个节点满足的条件 + var curChar = word[charIndex + 1]; + if (curChar == '.') + { + foreach (var nodeItem in node.m_nextNode) + { + var isOk = false; + if (nodeItem != null) + isOk = RecurSive(nodeItem, word, charIndex + 1); + + if (isOk) return true; + } + } + else + { + var cIndex = curChar - 'a'; + if (node.m_nextNode[cIndex] != null) + return RecurSive(node.m_nextNode[cIndex], word, charIndex + 1); + } + + return false; + } + } + + /** + * Your WordDictionary object will be instantiated and called as such: + * WordDictionary obj = new WordDictionary(); + * obj.AddWord(word); + * bool param_2 = obj.Search(word); + */ + } +} diff --git a/Week_04/id_9/LeetCode_720_9.cs b/Week_04/id_9/LeetCode_720_9.cs new file mode 100644 index 00000000..b1167f1f --- /dev/null +++ b/Week_04/id_9/LeetCode_720_9.cs @@ -0,0 +1,123 @@ +using System; +using System.Collections.Generic; +using System.Linq; +using System.Text; +using System.Threading.Tasks; + +namespace ProblemSolutions +{ + public class Problem720 : IProblem + { + public void RunProblem() + { + string[] words = new string[] { "a", "apple", "b", "be", "bee", "beer" }; + var temp = LongestWord(words); + } + + public string LongestWord(string[] words) + { + /* + * 实现思路： + * 1.使用数据源来“喂养”TrieTree + * 2.再让TrieTree告知我们，满足题目需求的结果 + * + * 时间复杂度：输入的单词，是要遍历一次的，构建trieTree，然后要拿到结果，还是要做回溯的，那么是对树所有节点又遍历一次 + * 空间复杂度：主要是TrieTree占用的存储空间 + * + * 使用此种方式，其实就是数据源越大越是有利 + */ + + var trieTree = new TrieTree(); + + foreach (var wordItem in words) + trieTree.AddWord(wordItem); + + return trieTree.GetLongestWord(); + } + + public class TrieTree + { + public class TrieNode + { + public string m_words; + public TrieNode[] m_nextIndex = new TrieNode[26]; + public bool m_isWord; + } + + private TrieNode m_rootNode = new TrieNode(); + + public void AddWord(string word) + { + var trieNodeTemp = m_rootNode; + foreach (var wordItem in word) + { + var intTemp = wordItem - 'a'; + + if (trieNodeTemp.m_nextIndex[intTemp] == null) + trieNodeTemp.m_nextIndex[intTemp] = new TrieNode(); + + trieNodeTemp = trieNodeTemp.m_nextIndex[intTemp]; + } + trieNodeTemp.m_isWord = true; + trieNodeTemp.m_words = word; + } + + public string GetLongestWord() + { + Recursive(m_rootNode, -1); + return longestWord; + } + + private int longest = 0; + private string longestWord = ""; + private void Recursive(TrieNode node, int length) + { + length++; + + if (node.m_isWord && length > longest) + { + longest = length; + longestWord = node.m_words; + } + + for (int i = 0; i < node.m_nextIndex.Length; i++) + { + if (node.m_nextIndex[i] != null && node.m_nextIndex[i].m_isWord) + { + Recursive(node.m_nextIndex[i], length); + } + } + } + } + + public string LongestWord1(string[] words) + { + /* + * 思路： + * 1.结果要求按照字典序来处理相同长度的单词，那么就先对数组做字典序排序，复杂度：O(nlogn); + * 2.遍历新的数组，寻找满足条件的最长单词 + * + * 时间复杂度：O(nlogn + n)，但是还要考虑到字符串比较的复杂度才行，字符串比较，应该是使用简单的比较方法才对 + * 空间复杂度：O(n)，因为会把所有的单词存入到set中去 + */ + + var forReturn = ""; + HashSet sets = new HashSet(); + + var sortedArray = words.OrderBy(i => i); + + foreach (var arrayItem in sortedArray) + { + if (arrayItem.Length == 1 || sets.Contains(arrayItem.Substring(0, arrayItem.Length - 1))) + { + if (forReturn.Length < arrayItem.Length) + forReturn = arrayItem; + + sets.Add(arrayItem); + } + } + + return forReturn; + } + } +} diff --git a/Week_04/id_9/NOTE.md b/Week_04/id_9/NOTE.md index c684e62f..0190ff24 100644 --- a/Week_04/id_9/NOTE.md +++ b/Week_04/id_9/NOTE.md @@ -1 +1,7 @@ -# 学习笔记 \ No newline at end of file +# 学习笔记 +1. 本次主要是练习TrieTree相关的题目，一个简单的一个中等的； +2. 两个题，其实都用到的是：TrieTree的构建+回溯法 +3. 当需要尝试各种可能性时，使用回溯法是很不错的； +4. 第1题，要找出满足条件的最长字符串，不遍历一遍TrieTree，是不会知道最长的字符串的 +5. 第2题，相当于做的是模糊匹配，有多种可能的匹配方式，只要一种匹配尝试成功了就是成功 +6. TrieTree是很实用的，构建也是相当简单的，单个的Trie节点，可以携带业务信息来加速查找，比如“是否是单词”，“当前字符是什么”，“若是一个单词，那么这个单词是什么” \ No newline at end of file