Leetcode 刷题必须Review 十七 Lintcode（423 492 541 421 575）
文章目录
- 423 · 有效的括号序列
- 492 · 队列维护
- 541 · 左旋右旋迭代器 II
- 421 · 简化路径
- 575 · 字符串解码
423 · 有效的括号序列
给定一个字符串所表示的括号序列，包含以下字符： ‘(’, ‘)’, ‘{’, ‘}’, ‘[’ and ‘]’， 判定是否是有效的括号序列。
括号必须依照 “()” 顺序表示， “()[]{}” 是有效的括号，但 “([)]” 则是无效的括号。
def isValidParentheses(self, s):
# write your code here
if not s: return True
stack = []
for i in s:
if i == '(' or i == '[' or i == '{':
stack.append(i)
else:
if not stack: return False
ele = stack.pop()
if (i == ')' and ele == '(') or (i == ']' and ele == '[') or (i == '}' and ele == '{'):
continue
else:
return False
return not stack
492 · 队列维护
按链表实现队列。支持以下基本方法：
enqueue(item).将新元素放入队列中。
dequeue(). 将第一个元素移出队列，返回它。
看了之前写的改了一下
class MyQueue:
"""
@param: item: An integer
@return: nothing
"""
def __init__(self):
self.queue = None
def enqueue(self, item):
# write your code here
if not self.queue:
self.queue = ListNode(item)
self.dummy = ListNode(-1, self.queue)
else:
self.queue.next = ListNode(item)
self.queue = self.queue.next
"""
@return: An integer
"""
def dequeue(self):
# write your code here
print(self.dummy.next.val)
res = self.dummy.next.val
self.dummy = self.dummy.next
return res
下面是我之前写的：
class MyQueue:
"""
@param: item: An integer
@return: nothing
"""
def __init__(self):
self.beforehead = self.tail = ListNode(-1)
def enqueue(self, item):
# write your code here
self.tail.next = ListNode(item)
self.tail = self.tail.next
"""
@return: An integer
"""
def dequeue(self):
# write your code here
if self.beforehead.next is None:
return -1
head_val = self.beforehead.next.val
self.beforehead = self.beforehead.next
return head_val
541 · 左旋右旋迭代器 II
和题目 左旋右旋迭代器 类似，在本题中，你将得到一个列表vecs，其中包括 k 个一维向量。
你的任务是通过 next 函数一个个地返回向量中的元素，按照 vecs[0][0], vecs[1][0]… vecs[k - 1][0], vecs[0][1], vecs[1][1]… vecs[k - 1][1], vecs[0][2], vecs[1][2]… vecs[k - 1][2]… 的顺序进行迭代。
没写出来。
from collections import deque
class ZigzagIterator2:
"""
@param: vecs: a list of 1d vectors
"""
def __init__(self, vecs):
# do intialization if necessary
max_col = max([len(vec) for vec in vecs])
self.queue = deque()
for j in range(max_col):
for i in range(len(vecs)):
try:
self.queue.append(vecs[i][j])
except IndexError:
pass
"""
@return: An integer
"""
def _next(self):
# write your code here
self.queue.popleft()
"""
@return: True if has next
"""
def hasNext(self):
# write your code here
if self.queue:
return True
# Your ZigzagIterator2 object will be instantiated and called as such:
# solution, result = ZigzagIterator2(vecs), []
# while solution.hasNext(): result.append(solution.next())
# Output result
看了之前写的，果然忘记了
import collections
class ZigzagIterator2:
"""
@param: vecs: a list of 1d vectors
"""
def __init__(self, vecs):
# do intialization if necessary
self.queue = collections.deque()
for vec in vecs:
if len(vec) > 0:
self.queue.append([iter(vec), len(vec)])
"""
@return: An integer
"""
def _next(self):
# write your code here
vec_iter, vec_len = self.queue.popleft()
value = next(vec_iter)
vec_len -= 1
if vec_len > 0:
self.queue.append([vec_iter, vec_len])
return value
"""
@return: True if has next
"""
def hasNext(self):
# write your code here
if self.queue:
return True
# Your ZigzagIterator2 object will be instantiated and called as such:
# solution, result = ZigzagIterator2(vecs), []
# while solution.hasNext(): result.append(solution.next())
# Output result
421 · 简化路径
给定一个文件的绝对路径(Unix-style)，请进行路径简化。
Unix中, . 表示当前目录, … 表示父目录。
结果必须以 / 开头，并且两个目录名之间有且只有一个 /。最后一个目录名(如果存在)后不能出现 / 。你需要保证结果是正确表示路径的最短的字符串。
有一个过不去：
class Solution:
"""
@param path: the original path
@return: the simplified path
"""
def simplify_path(self, path: str) -> str:
# write your code here
stack = []
li = path.split('/')
for i in li:
if i.isalnum():
stack.append(i)
elif i == ".":
continue
elif i == "..":
if stack: stack.pop()
return "/" + "/".join(stack)
下面是官方答案：
def simplifyPath(self, path):
path = path.split('/')
stack = []
for i in path:
if i == '..':
if len(stack):
stack.pop()
elif i != '.' and i != '':
stack.append(i)
return '/' + '/'.join(stack)
按照答案，改了下代码：
def simplify_path(self, path: str) -> str:
# write your code here
stack = []
li = path.split('/')
for i in li:
if i == "..":
if stack:
stack.pop()
elif i != "." and i != "":
stack.append(i)
return '/' + '/'.join(stack)
575 · 字符串解码
给出一个表达式 s，此表达式包括数字，字母以及方括号。在方括号前的数字表示方括号内容的重复次数(括号内的内容可以是字符串或另一个表达式)，请将这个表达式展开成一个字符串。
不会。直接看之前写的吧。
下面是之前写的。看了半天才想起来怎么做的。
def expressionExpand(self, s):
# write your code here
stack = []
number = 0
for ch in s:
if ch.isdigit():
number = number * 10 + int(ch)
elif ch == '[':
stack.append(number)
number = 0
elif ch == ']':
strs = []
while stack and not isinstance(stack[-1], int):
strs.append(stack.pop())
strs.reverse()
repeat = stack.pop()
for _ in range(repeat):
stack.append("".join(strs))
else:
stack.append(ch)
strs = []
while stack:
strs.append(stack.pop())
strs.reverse()
return ''.join(strs)
又自己写了一遍。学习了方法isinstance
def expressionExpand(self, s):
# write your code here
stack = []
number = 0
for ch in s:
if ch.isdigit():
number = number * 10 + int(ch)
elif ch == '[':
stack.append(number)
number = 0
elif ch == ']':
strs = []
while stack and not isinstance(stack[-1], int):
strs.append(stack.pop())
strs.reverse()
repeat = stack.pop()
for _ in range(repeat):
stack.append(''.join(strs))
else:
stack.append(ch)
ss = []
while stack:
ss.append(stack.pop())
ss.reverse()
return ''.join(ss)