题目描述

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

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 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

思路

深度优先遍历（dfs）+ 回溯

需要用到一个mark数组来标记是否走过

class Solution(object):def __init__(self):self.direction = [(0,1),(-1,0),(1,0),(0,-1)]def backtrack(self,i,j,board,word,mark):if len(word)==0:return Truefor dir in self.direction:cur_i = i+dir[0]cur_j = j+dir[1]if cur_i>=0 and cur_i<len(board) and cur_j>=0 and cur_j<len(board[0]) and board[cur_i][cur_j]==word[0]:if mark[cur_i][cur_j]==1: # 如果已经标记为使用过，则continuecontinuemark[cur_i][cur_j]=1if self.backtrack(cur_i,cur_j,board,word[1:],mark)==True:return Trueelse:mark[cur_i][cur_j]=0return Falsedef exist(self, board, word):""":type board: List[List[str]]:type word: str:rtype: bool"""rows = len(board)cols = len(board[0])mark = [[0]*cols for _ in range(rows)]for i in range(rows):for j in range(cols):if board[i][j] == word[0]:mark[i][j] = 1if self.backtrack(i,j,board,word[1:],mark)==True:return Trueelse:mark[i][j]=0return Falseif __name__ == "__main__":s=Solution()board = [["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]]word = "ABCCED"print(s.exist(board, word))