leetcode-79. 单词搜索

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 True
        for 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continue
                    continue
                mark[cur_i][cur_j]=1
                if self.backtrack(cur_i,cur_j,board,word[1:],mark)==True:
                    return True
                else:
                    mark[cur_i][cur_j]=0
        return False

    def 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] = 1
                    if self.backtrack(i,j,board,word[1:],mark)==True:
                        return True
                    else:
                        mark[i][j]=0
        return False

if __name__ == "__main__":
    s=Solution()
    board = [["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]]
    word = "ABCCED"
    print(s.exist(board, word))

原文地址：https://blog.csdn.net/m0_38098373/article/details/140626483

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