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…

一.题目说明 题目79. Word Search,给定一个由字符组成的矩阵,从矩阵中查找一个字符串是否存在.可以连续横.纵找.不能重复使用,难度是Medium. 二.我的解答 惭愧,我写了很久总是有问题,就先看正确的写法,下面是回溯法的代码: class Solution { public: int m,n; //左 上 右 下 int dx[4] = {-1,0,1,0}; int dy[4] = {0,1,0,-1}; bool exist(vector<vector<char>&g…

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…

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 m…

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…

题目 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…

［抄题］: 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…

使用的别人的思路,用一个二维数组记录每一个位置是否用过,然后通过递归来判断每一个位置是否符合 public class Solution { public boolean exist(char[][] board, String word) { if(board.length == 0) return false; int leni = board.length; int lenj = board[0].length; boolean[][] isVisited = new boolean[le…

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 >= &&…