(Java) LeetCode 79. Word Search —— 单词搜索 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 b...
翻译: 给定一个2D的字符数组和一个单词,找出这个词是否存在于网格中。 这个词可以用顺序相邻的细胞的字母来构造,在那里“相邻”的细胞是水平的或垂直的相邻的。同一个字母单元可能不止一次使用。 Example: board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] Giv...
search(board, visit, trie, i, j, new StringBuilder(), res); return new ArrayList<>(res); } private void search(char[][] board,boolean[][] visited,Trie trie,int i,int j, StringBuilder sb, Set<String> res) { if(i<0 || i>board.length-1 || j<0 || j>board[0].length-1 |...
https://leetcode.com/problems/word-search-ii/ 最直观的思路就是对于每个word,都对board进行一次dfs搜索。这样在words太多的情况下会TLE. 这里考虑到board上有些point,对于任意word或者任意word的后缀都没有再继续搜索的必要,所以这里采用trie字典树存储所有带搜索的word。这样就只用dfs board一遍,就可以把所有的在...
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1. Description Word Search II 2. Solution class Solution{public:vector<string>findWords(vector<vector<char>>&board,vector<string>&words){vector<string>result;introws=board.size();intcolumns=board[0].size();unordered_set<string>s;for(string word:words){s.insert(word);}for(string word:s){...
Java: class Solution { public boolean exist(char[][] board, String word) { for (int i = 0; i < board.length; i++) for (int j = 0; j < board[0].length; j++) if (board[i][j] == word.charAt(0)) if (backtrack(board, word, i, j, board.length, board[0].length, 0...
public boolean search(String word) { return helper(word, 0, root); } private boolean helper(String word, int start, TrieNode node) { if (start == word.length()) return node.isWord; char ch = word.charAt(start); if (ch == '.') { ...
Leetcode 212 Word Search II的字典树构建过程是怎样的? Given a 2D board and a list of words from the dictionary, find all words in the board. Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The ...
public class Solution { public boolean exist(char[][] board, String word) { if(word == null || word.length() == 0) return true; else if(board == null || board.length == 0) return false; boolean[][] isVisited = new boolean[board.length][board[0].length]; ...