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 us…

本文始发于个人公众号:TechFlow,原创不易,求个关注 今天是LeetCode专题第48篇文章,我们一起来看看LeetCode当中的第79题,搜索单词(Word Search). 这一题官方给的难度是Medium,通过率是34.5%,点赞3488,反对170.单从这份数据上来看,这题的质量很高,并且难度比之前的题目稍稍大一些.我个人觉得通过率是比官方给的题目难得更有参考意义的指标,10%到20%可以认为是较难的题,30%左右是偏难的题.50%是偏易题,所以如果看到某题标着Hard,但是通过率…

https://leetcode.com/problems/word-search/ class Solution { public: struct Trie{ Trie *next[]; bool isWord; Trie() { this->isWord = false; for(auto &c: next) c = NULL; } }; void insert(Trie *root, string word) { Trie *p = root; for(auto &c: wor…

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 us…

题目链接:https://leetcode.com/problems/word-search/#/description 给出一个二维字符表,并给出一个String类型的单词,查找该单词是否出现在该二维字符表中. For example,Given board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] word = "ABCCED", -> returns true,word = "SEE…

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 us…

https://www.cnblogs.com/grandyang/p/4332313.html 在一个矩阵中能不能找到string的一条路径 这个题使用的是dfs.但这个题与number of islands有点不同,那个题中visited过的就不用再扫了,但是这个需要进行回溯回来. 所以使用了visited[i][j] = false; 本质区别还是那个题是找一片区域,这个题更像是找一条路径. 错误一:如果board不引用,会报超内存 错误二: 这个写法和正确写法基本相同,但错误写法需要每次…

79. 单词搜索 给定一个二维网格和一个单词,找出该单词是否存在于网格中. 单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中"相邻"单元格是那些水平相邻或垂直相邻的单元格.同一个单元格内的字母不允许被重复使用. 示例: board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] 给定 word = "ABCCED", 返回 true. 给定 word = "SEE"…

题目 给定一个二维网格和一个单词,找出该单词是否存在于网格中. 单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中"相邻"单元格是那些水平相邻或垂直相邻的单元格.同一个单元格内的字母不允许被重复使用. 示例: board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] 给定 word = "ABCCED", 返回 true. 给定 word = "SEE", 返回…

原题地址 依次枚举起始点,DFS+回溯 代码: bool dfs(vector<vector<char> > &board, int r, int c, string word) { int m = board.size(); ].size(); ][] = {{-, }, {, }, {, -}, {, }}; if (word.empty()) return true; && r < m && c >= &&…