[leetcode trie]212. Word Search II
Given a 2D board and a list of words from the dictionary, find all words in the board.
Each word must 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 in a word.
For example,
Given words = ["oath","pea","eat","rain"] and board =
[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]
Return ["eat","oath"].
题意:查找哪些单词在二维字母数组中出现
思路:1.用所有的单词建立字典树
2.遍历二维数组中的所有位置,以这个位置开头向二维数组上下左右方向扩展的字符串是否在字典树中出现
评论区还有一种用复数和字典树的解题方法,太帅了
class Solution(object):
def findWords(self, board, words):
trie = {}
for w in words:
cur = trie
for c in w:
cur = cur.setdefault(c,{})
cur[None] = None
self.flag = [[False]*len(board[0]) for _ in range(len(board))]
self.res = set()
for i in range(len(board)):
for j in range(len(board[0])):
self.find(board,trie,i,j,'')
return list(self.res) def find(self,board,trie,i,j,str):
if None in trie:
self.res.add(str)
if i<0 or i>=len(board) or j<0 or j>=len(board[0]):
return
if not self.flag[i][j] and board[i][j] in trie:
self.flag[i][j] = True
self.find(board,trie[board[i][j]],i-1,j,str+board[i][j])
self.find(board,trie[board[i][j]],i+1,j,str+board[i][j])
self.find(board,trie[board[i][j]],i,j+1,str+board[i][j])
self.find(board,trie[board[i][j]],i,j-1,str+board[i][j])
self.flag[i][j] = False
return
[leetcode trie]212. Word Search II的更多相关文章
- 【leetcode】212. Word Search II
Given an m x n board of characters and a list of strings words, return all words on the board. Each ...
- 【LeetCode】212. Word Search II 解题报告(C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 前缀树 日期 题目地址:https://leetco ...
- leetcode 79. Word Search 、212. Word Search II
https://www.cnblogs.com/grandyang/p/4332313.html 在一个矩阵中能不能找到string的一条路径 这个题使用的是dfs.但这个题与number of is ...
- [LeetCode] 212. Word Search II 词语搜索之二
Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...
- Java for LeetCode 212 Word Search II
Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...
- [LeetCode] 212. Word Search II 词语搜索 II
Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...
- 212. Word Search II
题目: Given a 2D board and a list of words from the dictionary, find all words in the board. Each word ...
- [LeetCode#212]Word Search II
Problem: Given a 2D board and a list of words from the dictionary, find all words in the board. Each ...
- 79. 212. Word Search *HARD* -- 字符矩阵中查找单词
79. Word Search Given a 2D board and a word, find if the word exists in the grid. The word can be co ...
随机推荐
- 【CodeForces】700 D. Huffman Coding on Segment 哈夫曼树+莫队+分块
[题目]D. Huffman Coding on Segment [题意]给定n个数字,m次询问区间[l,r]的数字的哈夫曼编码总长.1<=n,m,ai<=10^5. [算法]哈夫曼树+莫 ...
- linux源码安装 rpm命令
安装dhcp为例: 挂载光盘文件到/media目录 #mount /dev/sr0 /media 打开/media目录下的Packages目录 #cd /media/Packages 查看系统是否安装 ...
- 2017ACM暑期多校联合训练 - Team 9 1010 HDU 6170 Two strings (dp)
题目链接 Problem Description Giving two strings and you should judge if they are matched. The first stri ...
- c# 超长字符串截取固定长度后显示...(超长后面显示点点点) 通用方法
通用方法: 此方法是采用unicode编码方式,一个汉字为2个字节,一个数字or字母是1个字节,此方法传入的第二个长度参数是unicode长度. 所以不用考虑截取的字符串是汉字还是英文字母的问题,参数 ...
- 【日记】NOIP2018
day-2: 最后一次走出机房,刚下过几天的雨,感受到的是彻骨的寒意.下午离开教室,跟班主任请了接下来几天的假,班主任斜视了我一眼,哼了一声,确认了一下,不再理会我了.班里的同学或是忙着自己的作业,或 ...
- mariadb/mysql使用Navicat连接报错
[问题1] 使用Navicat连接服务器的mariadb/mysql时报错 access denied for user root@192.168.xx.xx(using password:yes) ...
- MongoDB安全:创建角色(User-Defined Roles)
MongoDB已经定义了一些内建角色,同时还提供了用户自定义角色的功能,以满足用户千差万别的需求. 官文User-Defined Roles中对其有简略介绍,但要熟悉怎么创建角色,还需要了解下面的这些 ...
- R语言以及RStdio的安装
R语言: 首先从官网上下载R安装包, 提供了Linux, (Mac) OS X, Windows的安装包相关下载链接. RStdio: RStdio(官网)是R言语非常实用的IDE, 是一个免费的软件 ...
- Java---容器基础总结
Java提供了大量持有对象的方式: (1) 数组将数字与对象联系起来. 它保存类型明确的对象,查询对象时,不需要对结果做类型转换.它可以是多维的, 可以保存基本类型的数据. 但是,数组一旦生成,其容量 ...
- 20155309 《Java程序设计》实验三(Java面向对象程序设计)实验报告
一.实验内容及步骤 (一)编码标准 在IDEA中使用工具(Code->Reformate Code)把代码重新格式化. (二)在码云上把自己的学习搭档加入自己的项目中,确认搭档的项目加入自己后, ...