Python每日一练(20230327)

目录

1. 最大矩形  🌟🌟🌟

2. 反转链表 II  🌟🌟

3. 单词接龙 II  🌟🌟🌟

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1. 最大矩形

给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。

示例 1：

输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：6
解释：最大矩形如上图所示。

示例 2：

输入：matrix = []
输出：0

示例 3：

输入：matrix = [["0"]]
输出：0

示例 4：

输入：matrix = [["1"]]
输出：1

示例 5：

输入：matrix = [["0","0"]]
输出：0

提示：

• rows == matrix.length
• cols == matrix[0].length
• 0 <= row, cols <= 200
• matrix[i][j] 为 '0' 或 '1'

出处：

https://edu.csdn.net/practice/23885007

代码：

from typing import Listclass Solution(object):def maximalRectangle(self, matrix):""":type matrix: List[List[str]]:rtype: int"""if matrix is None or len(matrix) == 0:return 0ls_row, ls_col = len(matrix), len(matrix[0])left, right, height = [0] * ls_col, [ls_col] * ls_col, [0] * ls_colmaxA = 0for i in range(ls_row):curr_left, curr_right = 0, ls_colfor j in range(ls_col):if matrix[i][j] == '1':height[j] += 1else:height[j] = 0for j in range(ls_col):if matrix[i][j] == '1':left[j] = max(left[j], curr_left)else:left[j], curr_left = 0, j + 1for j in range(ls_col - 1, -1, -1):if matrix[i][j] == '1':right[j] = min(right[j], curr_right)else:right[j], curr_right = ls_col, jfor j in range(ls_col):maxA = max(maxA, (right[j] - left[j]) * height[j])return maxA# %%
s = Solution()
matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
print(s.maximalRectangle(matrix))
matrix = []
print(s.maximalRectangle(matrix))
matrix = [["0"]]
print(s.maximalRectangle(matrix))
matrix = [["1"]]
print(s.maximalRectangle(matrix))
matrix = [["0","0"]]
print(s.maximalRectangle(matrix))

输出：

6
0
0
1
0

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2. 反转链表 II

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

示例 1：

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]

示例 2：

输入：head = [5], left = 1, right = 1
输出：[5]

提示：

• 链表中节点数目为 n
• 1 <= n <= 500
• -500 <= Node.val <= 500
• 1 <= left <= right <= n

进阶： 你可以使用一趟扫描完成反转吗？

出处：

https://edu.csdn.net/practice/23885008

代码：

class ListNode(object):def __init__(self, x):self.val = xself.next = Noneclass LinkList:def __init__(self):self.head=Nonedef initList(self, data):self.head = ListNode(data[0])r=self.headp = self.headfor i in data[1:]:node = ListNode(i)p.next = nodep = p.nextreturn rdef convert_list(self,head):ret = []if head == None:returnnode = headwhile node != None:ret.append(node.val)node = node.nextreturn retclass Solution(object):def reverseBetween(self, head, m, n):""":type head: ListNode:type m: int:type n: int:rtype: ListNode"""if m == n:return headsplit_node, prev, curr = None, None, headcount = 1while count <= m and curr is not None:if count == m:split_node = prevprev = currcurr = curr.nextcount += 1tail, next_node = prev, Nonewhile curr is not None and count <= n:next_temp = curr.nextcurr.next = prevprev = currcurr = next_tempcount += 1if split_node is not None:split_node.next = previf tail is not None:tail.next = currif m == 1:return prevreturn head# %%
l = LinkList()
list1 = [1,2,3,4,5]
l1 = l.initList(list1)
left = 2
right = 4
s = Solution()
print(l.convert_list(s.reverseBetween(l1, left, right)))

输出：

[1, 4, 3, 2, 5]

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3. 单词接龙 II

按字典 wordList 完成从单词 beginWord 到单词 endWord 转化，一个表示此过程的 转换序列 是形式上像 beginWord -> s1 -> s2 -> ... -> sk 这样的单词序列，并满足：

• 每对相邻的单词之间仅有单个字母不同。
• 转换过程中的每个单词 si（1 <= i <= k）必须是字典 wordList 中的单词。注意，beginWord 不必是字典 wordList 中的单词。
• sk == endWord

给你两个单词 beginWord 和 endWord ，以及一个字典 wordList 。请你找出并返回所有从 beginWord 到 endWord 的 最短转换序列 ，如果不存在这样的转换序列，返回一个空列表。每个序列都应该以单词列表 [beginWord, s1, s2, ..., sk] 的形式返回。

示例 1：

输入：beginWord = "hit", endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]
输出：[["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
解释：存在 2 种最短的转换序列：
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"

示例 2：

输入：beginWord = "hit", endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
输出：[]
解释：endWord "cog" 不在字典 wordList 中，所以不存在符合要求的转换序列。

提示：

• 1 <= beginWord.length <= 7
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• beginWord、endWord 和 wordList[i] 由小写英文字母组成
• beginWord != endWord
• wordList 中的所有单词 互不相同

出处：

https://edu.csdn.net/practice/23885009

代码：

from typing import Listclass Solution:def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:import collectionsans = []steps = float("inf")if not beginWord or not endWord or not wordList or endWord not in wordList:return []word_dict = collections.defaultdict(list)L = len(beginWord)for word in wordList:for i in range(L):word_dict[word[:i] + "*" + word[i+1:]].append(word)queue = [[beginWord]]cost = {beginWord : 0}while queue:words_list= queue.pop(0)cur_word = words_list[-1]if cur_word == endWord:ans.append(words_list[:])else:for i in range(L):intermediate_word = cur_word[:i] + "*" + cur_word[i+1:]for word in word_dict[intermediate_word]:w_l_temp = words_list[:]if word not in cost or cost[word] >= cost[cur_word] + 1:cost[word] = cost[cur_word] + 1w_l_temp.append(word)queue.append(w_l_temp[:])return ans# %%
s = Solution()
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]
print(s.findLadders(beginWord, endWord, wordList))

输出：

[['hit', 'hot', 'dot', 'dog', 'cog'], ['hit', 'hot', 'lot', 'log', 'cog']]

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