LeetCode解题笔记79 Word Search

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 used more than once.

For example,

Given board =

[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

word = 

"ABCCED"

,
-> returns 

true

,
word = 

"SEE"

,
-> returns 

true

,
word = 

"ABCB"

,
-> returns 

false

.
python解法一：

采用DFS算法

class Solution:
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
l = len(word)
m = len(board)
n = len(board[0])
def dfs(i,j,k):
if board[i][j]!=word[k]:
return False
if k+1 == l:
return True
board[i][j]+="#"
has = (i>=1 and dfs(i-1,j,k+1)) or (i<m-1 and dfs(i+1,j,k+1)) or (j>=1 and dfs(i,j-1,k+1)) or (j<n-1 and dfs(i,j+1,k+1))
board[i][j] = board[i][j][0]
return has
for i in range(m):
for j in range(n):
if dfs(i,j,0):
return True
return False