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.lengthcols == matrix[0].length0 <= row, cols <= 200matrix[i][j]为'0'或'1'
出处：
https://edu.csdn.net/practice/23885007
代码：
from typing import List
class Solution(object):
def maximalRectangle(self, matrix):
"""
:type matrix: List[List[str]]
:rtype: int
"""
if matrix is None or len(matrix) == 0:
return 0
ls_row, ls_col = len(matrix), len(matrix[0])
left, right, height = [0] * ls_col, [ls_col] * ls_col, [0] * ls_col
maxA = 0
for i in range(ls_row):
curr_left, curr_right = 0, ls_col
for j in range(ls_col):
if matrix[i][j] == '1':
height[j] += 1
else:
height[j] = 0
for j in range(ls_col):
if matrix[i][j] == '1':
left[j] = max(left[j], curr_left)
else:
left[j], curr_left = 0, j + 1
for 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, j
for 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
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 <= 5001 <= 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 = x
self.next = None
class LinkList:
def __init__(self):
self.head=None
def initList(self, data):
self.head = ListNode(data[0])
r=self.head
p = self.head
for i in data[1:]:
node = ListNode(i)
p.next = node
p = p.next
return r
def convert_list(self,head):
ret = []
if head == None:
return
node = head
while node != None:
ret.append(node.val)
node = node.next
return ret
class Solution(object):
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
if m == n:
return head
split_node, prev, curr = None, None, head
count = 1
while count <= m and curr is not None:
if count == m:
split_node = prev
prev = curr
curr = curr.next
count += 1
tail, next_node = prev, None
while curr is not None and count <= n:
next_temp = curr.next
curr.next = prev
prev = curr
curr = next_temp
count += 1
if split_node is not None:
split_node.next = prev
if tail is not None:
tail.next = curr
if m == 1:
return prev
return 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]
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 <= 7endWord.length == beginWord.length1 <= wordList.length <= 5000wordList[i].length == beginWord.lengthbeginWord、endWord和wordList[i]由小写英文字母组成beginWord != endWordwordList中的所有单词 互不相同
出处：
https://edu.csdn.net/practice/23885009
代码：
from typing import List
class Solution:
def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
import collections
ans = []
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] + 1
w_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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