模拟卷Leetcode【普通】079. 单词搜索
作者:互联网
079. 单词搜索
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
提示:
m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board 和 word 仅由大小写英文字母组成
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/word-search
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
代码:
import time
from typing import List
class Solution:
def __init__(self):
pass
def dfs(self, visited,row_id,col_id,word_id):
if row_id<0 or row_id>=self.height or \
col_id<0 or col_id>=self.width or \
self.word[word_id]!=self.board[row_id][col_id] or \
visited[row_id][col_id]:
return False
elif word_id==len(self.word)-1:
return True
else:
visited[row_id][col_id]=True
res = self.dfs(visited, row_id + 1, col_id, word_id+1) or \
self.dfs(visited, row_id - 1, col_id, word_id + 1) or \
self.dfs(visited, row_id, col_id + 1, word_id + 1) or \
self.dfs(visited, row_id, col_id - 1, word_id + 1)
visited[row_id][col_id]=False
return res
def exist(self, board: List[List[str]], word: str) -> bool:
if len(board)==0 or len(board[0])==0 :return False
self.board = board
self.height = len(board)
self.width = len(board[0])
self.word = word
visited=[[0 for _ in range(len(board[0]))] for _ in range(len(board))]
for row_id,row in enumerate(board):
for col_id,item in enumerate(row):
if self.dfs(visited,row_id,col_id,0):
return True
return False
def test(data_test):
s = Solution()
return s.exist(*data_test)
def test_obj(data_test):
result = [None]
obj = Solution(*data_test[1][0])
for fun, data in zip(data_test[0][1::], data_test[1][1::]):
if data:
res = obj.__getattribute__(fun)(*data)
else:
res = obj.__getattribute__(fun)()
result.append(res)
return result
if __name__ == '__main__':
datas = [
[],
]
for data_test in datas:
t0 = time.time()
print('-' * 50)
print('input:', data_test)
print('output:', test(data_test))
print(f'use time:{time.time() - t0}s')
备注:
GitHub:https://github.com/monijuan/leetcode_python
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leetcode_python.utils详见汇总页说明
先刷的题,之后用脚本生成的blog,如果有错请留言,我看到了会修改的!谢谢!
标签:word,self,Leetcode,单词,col,board,079,id,row 来源: https://blog.csdn.net/qq_34451909/article/details/121602773