Python每日一练(20230416)
目录
1. 有效数字 🌟🌟🌟
2. 二叉树的最大深度 🌟
3. 单词搜索 🌟🌟
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1. 有效数字
有效数字(按顺序)可以分成以下几个部分:
- 一个 小数 或者 整数
- (可选)一个
'e'
或'E'
,后面跟着一个 整数
小数(按顺序)可以分成以下几个部分:
- (可选)一个符号字符(
'+'
或'-'
) - 下述格式之一:
- 至少一位数字,后面跟着一个点
'.'
- 至少一位数字,后面跟着一个点
'.'
,后面再跟着至少一位数字 - 一个点
'.'
,后面跟着至少一位数字
- 至少一位数字,后面跟着一个点
整数(按顺序)可以分成以下几个部分:
- (可选)一个符号字符(
'+'
或'-'
) - 至少一位数字
部分有效数字列举如下:
["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"]
部分无效数字列举如下:
["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"]
给你一个字符串 s
,如果 s
是一个 有效数字 ,请返回 true
。
示例 1:
输入:s = "0" 输出:true
示例 2:
输入:s = "e" 输出:false
示例 3:
输入:s = "." 输出:false
示例 4:
输入:s = ".1" 输出:true
提示:
1 <= s.length <= 20
s
仅含英文字母(大写和小写),数字(0-9
),加号'+'
,减号'-'
,或者点'.'
。
以下程序实现了这一功能,请你填补空白处内容:
```python
class Solution(object):
def isNumber(self, s):
s = s.strip()
ls, pos = len(s), 0
if ls == 0:
return False
if s[pos] == '+' or s[pos] == '-':
pos += 1
isNumeric = False
while pos < ls and s[pos].isdigit():
pos += 1
isNumeric = True
_____________________________;
elif pos < ls and s[pos] == 'e' and isNumeric:
isNumeric = False
pos += 1
if pos < ls and (s[pos] == '+' or s[pos] == '-'):
pos += 1
while pos < ls and s[pos].isdigit():
pos += 1
isNumeric = True
if pos == ls and isNumeric:
return True
return False
# %%
s = Solution()
print(s.isNumber(s = "0"))
```
出处:
https://edu.csdn.net/practice/25740860
代码:
class Solution(object):def isNumber(self, s):s = s.strip()ls, pos = len(s), 0if ls == 0:return Falseif s[pos] == '+' or s[pos] == '-':pos += 1isNumeric = Falsewhile pos < ls and s[pos].isdigit():pos += 1isNumeric = Trueif pos < ls and s[pos] == '.':pos += 1while pos < ls and s[pos].isdigit():pos += 1isNumeric = Trueelif pos < ls and s[pos] == 'e' and isNumeric:isNumeric = Falsepos += 1if pos < ls and (s[pos] == '+' or s[pos] == '-'):pos += 1while pos < ls and s[pos].isdigit():pos += 1isNumeric = Trueif pos == ls and isNumeric:return Truereturn False
# %%
s = Solution()
print(s.isNumber(s = "0"))
print(s.isNumber(s = "e"))
print(s.isNumber(s = "."))
print(s.isNumber(s = ".1"))
输出:
True
False
False
True
2. 二叉树的最大深度
给定一个二叉树,找出其最大深度。
二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
说明: 叶子节点是指没有子节点的节点。
示例:
给定二叉树 [3,9,20,null,null,15,7]
,
3/ \\9 20/ \\ 15 7
返回它的最大深度 3 。
出处:
https://edu.csdn.net/practice/25740861
代码:
class TreeNode:def __init__(self, x):self.val = xself.left = Noneself.right = Nonedef listToTree(lst: list) -> TreeNode:if not lst:return Noneroot = TreeNode(lst[0])queue = [root]i = 1while i < len(lst):node = queue.pop(0)if lst[i] is not None:node.left = TreeNode(lst[i])queue.append(node.left)i += 1if i < len(lst) and lst[i] is not None:node.right = TreeNode(lst[i])queue.append(node.right)i += 1return rootclass Solution:def maxDepth(self, root: TreeNode) -> int:if root is None:return 0maxdepth = max(self.maxDepth(root.left),self.maxDepth(root.right)) + 1return maxdepth
# %%
s = Solution()
null = None
nums = [3,9,20,null,null,15,7]
root = listToTree(nums)
print(s.maxDepth(root))
输出:
3
代码2:DFS
class TreeNode:def __init__(self, x):self.val = xself.left = Noneself.right = Nonedef listToTree(lst: list) -> TreeNode:if not lst:return Noneroot = TreeNode(lst[0])queue = [root]i = 1while i < len(lst):node = queue.pop(0)if lst[i] is not None:node.left = TreeNode(lst[i])queue.append(node.left)i += 1if i < len(lst) and lst[i] is not None:node.right = TreeNode(lst[i])queue.append(node.right)i += 1return rootclass Solution:def maxDepth(self, root: TreeNode):if not root:return 0depth = []def dfs(node, nodes=[]):if not node:returnnodes.append(node.val)if node.left or node.right:dfs(node.left, nodes)dfs(node.right, nodes)else:depth.append(len(nodes))nodes.pop()dfs(root)return max(depth)
# %%
s = Solution()
null = None
nums = [3,9,20,null,null,15,7]
root = listToTree(nums)
print(s.maxDepth(root))
3. 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
进阶:你可以使用搜索剪枝的技术来优化解决方案,使其在 board
更大的情况下可以更快解决问题?
出处:
https://edu.csdn.net/practice/25740862
代码:
class Solution(object):def exist(self, board, word):""":type board: List[List[str]]:type word: str:rtype: bool"""check_board = [[True] * len(board[0]) for _ in range(len(board))]for i in range(len(board)):for j in range(len(board[0])):if board[i][j] == word[0] and check_board:check_board[i][j] = Falseres = self.check_exist(check_board, board, word, 1, len(word), i, j)if res:return Truecheck_board[i][j] = Truereturn Falsedef check_exist(self, check_board, board, word, index, ls, row, col):if index == ls:return Truefor temp in [(0, 1),(0, -1),(1, 0),(-1, 0)]:curr_row = row + temp[0]curr_col = col + temp[1]if curr_row >= 0 and curr_row < len(board) and curr_col >= 0 and curr_col < len(board[0]):if check_board[curr_row][curr_col] and board[curr_row][curr_col] == word[index]:check_board[curr_row][curr_col] = Falseres = self.check_exist(check_board, board, word, index + 1, len(word), curr_row, curr_col)if res:return rescheck_board[curr_row][curr_col] = Truereturn False
if __name__ == "__main__":s = Solution()print (s.exist(board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"))
输出:
True
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