题:https://leetcode.com/problems/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.
Example:
board =
[
[‘A’,’B’,’C’,’E’],
[‘S’,’F’,’C’,’S’],
[‘A’,’D’,’E’,’E’]
]
Given word = “ABCCED”, return true.
Given word = “SEE”, return true.
Given word = “ABCB”, return false.
思路
总体就是递归遍历。
但有两个点需要注意:
1.遍历的时候,需要放置回溯。做法的技巧在于 原值修改,然后 递归,最后原值复原。
2.可能 board 与 word 都是一个值所以, if p == len(word): 与 if px == -1 or py == -1: 的位置需要注意
code
def get_up(px, py, board):
if px - 1 >= 0:
return px-1, py
return -1, -1
def get_down(px, py, board):
if px + 1 < len(board):
return px +1, py
return -1, -1
def get_left(px, py, board):
if py - 1 >= 0:
return px , py -1
return -1, -1
def get_right(px, py, board):
if py + 1 < len(board[0]):
return px , py + 1
return -1, -1
def is_word(px, py, p, board, word):
if p == len(word):
return True
if px == -1 or py == -1:
return False
# print(p,px,py)
if word[p] != board[px][py]:
return False
tmpchar = board[px][py]
board[px][py] = '-1'
bl_flag = is_word(get_up(px, py, board)[0], get_up(px, py, board)[1], p + 1, board, word) or is_word(
get_down(px, py, board)[0], get_down(px, py, board)[1], p + 1, board, word) or is_word(
get_left(px, py, board)[0], get_left(px, py, board)[1], p + 1, board, word) or is_word(
get_right(px, py, board)[0], get_right(px, py, board)[1], p + 1, board, word)
board[px][py] = tmpchar
return bl_flag
class Solution:
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
for i in range(len(board)):
for j in range(len(board[0])):
if is_word(i, j, 0, board, word):
return True
return False