LeetCode-Word Search 很简单的dfs; 1classSolution {2public:3boolexist(vector<vector<char> > &board,stringword) {4//Start typing your C/C++ solution below5//DO NOT write int main() function6if(word.empty()) {7returntrue;8}9if(board.empty()) {10returnfalse;11}12intm =board.size()...
class Solution { public: bool exist(vector<vector<char>>& board, string word) { h = board.size(); w = board[0].size(); for(int i = 0; i < h; i++){ for(int j = 0; j < w; ++j){ if(searchexist(board, word, 0, i, j)) return true; } } return false; } int se...
题目地址:https://leetcode.com/problems/word-search/description/ 题目描述 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 neigh...
leetcode 79. Word Search DFS 单词搜索 + 深度优先遍历,Givena2Dboardandaword,findifthewordexistsinthegrid.Thewordcanbec
/** @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"...
class Solution { public: bool exist(vector<vector<char>>& board, string word) { for (int i = 0; i < board.size(); i++) for (int j = 0; j < board[0].size(); j++) if (board[i][j] == word[0]) if (backtrack(board, word, i, j, board.size(), board[0].size()...
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...
clean codeMap<String, Integer>wordMap=newHashMap<>();// Using try-with-resource statement for automatic resource managementtry(FileInputStreamfis=newFileInputStream(fileName);DataInputStreamdis=newDataInputStream(fis);BufferedReaderbr=newBufferedReader(newInputStreamReader(dis))) {// words are ...
链接:https://leetcode.com/problems... 这道题可以用dp的方法,和word break一样,多加个循环,复杂度是O(N^3),这道题注意下,字典比较大,用第二种来写dp function会超时,只能用第一种。 public class Solution { public List<String> findAllConcatenatedWordsInADict(String[] words) { ...
leetcode 240. Search a 2D Matrix II 思路:从左下或者右上开始进行搜索,因为比如从左下开始搜索,若目标数大于此时的数,接下来只能向右搜,若小于,接着只能向上搜。若是从左上开始进行搜索,若目标数大于此时的数,会有两个方向可以走。 ...240. Search a 2D Matrix II class Solution { public: bool ...