Leetcode练习(Python)：数组类：第79题：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。
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。
思路：
深度优先和回溯法结合
程序1：
class Solution(object):
directs = [(0, 1), (0, -1), (1, 0), (-1, 0)]
def exist(self, board: List[List[str]], word: str) -> bool:
row = len(board)
if row == 0:
return False
column = len(board[0])
visited = [[0 for _ in range(column)] for _ in range(row)]
for index1 in range(len(board)):
for index2 in range(len(board[0])):
if board[index1][index2] == word[0]:
visited[index1][index2] = 1
if self.backtrack(index1, index2, visited, board, word[1:]) == True:
return True
else:
visited[index1][index2] = 0
return False
def backtrack(self, i, j, visited, board, word):
if len(word) == 0:
return True
for direct in self.directs:
cur_i = i + direct[0]
cur_j = j + direct[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 visited[cur_i][cur_j] == 1:
continue
visited[cur_i][cur_j] = 1
if self.backtrack(cur_i, cur_j, visited, board, word[1:]) == True:
return True
else:
visited[cur_i][cur_j] = 0
return False
程序2：代码更简洁
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
for index1 in range(len(board)):
for index2 in range(len(board[0])):
if self.dfs(board, index1, index2, word) == True:
return True
return False
def dfs(self, board, i, j, word):
if len(word) == 0:
return True
if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or board[i][j] != word[0]:
return False
tmp = board[i][j]
board[i][j] = '0'
if self.dfs(board, i + 1, j, word[1:]) or self.dfs(board, i - 1, j, word[1:]) \
or self.dfs(board, i, j + 1, word[1:]) or self.dfs(board, i, j - 1, word[1:]) == True:
return True
board[i][j] = tmp
return False
