1. Word Search Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once. For exam...
leetcode70word-search 题目描述给出一个二维字符数组和一个单词,判断单词是否在数组中出现,单词由相邻单元格的字母连接而成,相邻单元指的是上下左右相邻。同一单元格的字母不能多次使用。例如: 给出的字符数组= [↵ ["ABCE"],↵ ["SFCS"],↵ ["ADEE"]↵] 单词...
http://bookshadow.com/weblog/2015/05/19/leetcode-word-search-ii/ 这里的trie用child dict表示子节点,key是letter,value是node。我们要理解成每个node是没有value的,所以父节点到子节点的path才对应一个letter,即这里child dict的key。 借由下图,解释一下trie的delete操作。以前以为isWord只可能在叶子节点才会...
Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once. For example, Given boar...
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/** @lc app=leetcodeid=212 lang=cpp** [212] Word Search II** https://leetcode.com/problems/word-search-ii/description/** algorithms* Hard (29.77%)* Likes: 1405* Dislikes: 78* Total Accepted: 132.7K* Total Submissions: 440.4K* Testcase Example: '[["o","a","a","n"],["e"...
参考LeetCode #208 Implement Trie (Prefix Tree) 实现 Trie (前缀树)和LeetCode #79 Word Search 单词搜索 将words数组中的所有 word插入到前缀树中 再在board数组中进行 dfs搜索 时间复杂度O(mn * 3 ^ l), 空间复杂度O(lk), 其中 m和 n分别是 board数组的长宽, l是 words数组中的单词长度, k为 wo...
79 Word Search 单词搜索 Description: Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more...
LeetCode Question word search Deion: Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more...
LeetCode Question word search Deion: Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more...