Leetcode 79:Word Search 题目链接: https://leetcode.com/problems/word-search/description/ 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......
``` import java.util.ArrayList; import java.util.Arrays; import java.util.List; /** * Source : https://oj.leetcode.com/problems/word-search/ * * * Giv
leetcode---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. ...
Can you solve this real interview question? Word Search II - Given an m x n board of characters and a list of strings words, return all words on the board. Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells a
Word Search -- LeetCode 原题链接: http://oj.leetcode.com/problems/word-search/ 这道题很容易感觉出来是图的题目,其实本质上还是做深度优先搜索。基本思路就是从某一个元素出发,往...LeetCode - word-search 题目: Given a 2D board and a word, find if the word exists in the grid. The ...
visited[newx][newy] && searchWord(board, word, index + 1, newx, newy)) return true; } visited[startx][starty] = false; } return false; } public: bool exist(vector<vector<char>>& board, string word) { m = board.size(); assert(m > 0); n = board[0].size(); assert(n ...
leetcode 79. Word Search DFS 单词搜索 + 深度优先遍历,Givena2Dboardandaword,findifthewordexistsinthegrid.Thewordcanbec
中实现的字典树来保存words,然后修改word search I中的代码,在dfs中保存路径字符串,如果该字符串是一个单词(通过search字典树实现)则加入result,如果该字符串是字典树的一个prefix,且则继续往下(四个方向)进行搜索,且这四个方向对应的二维数组中的位置,不能在这之前已被访问(即同一个位置,在一个word的搜索过程...
*/ bool search(string word) { node* p = root; for(int i = 0; i < word.size(); i++) { if(!p->next[word[i] - 'a']) return false; p = p->next[word[i] - 'a']; } return p->flag == 1; } /** Returns if there is any word in the trie that starts with the ...
(board, visited, row + 1, col, word, idx) // 下 || search(board, visited, row, col - 1, word, idx); // 左 // 假设没有找到路径就回溯 if (!hasPath) { visited[row][col] = false; idx[0]--; } } return hasPath; } /** * 判定訪问的位置是否合法 * * @param board 字符...