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Update stack_queue.md
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‎data_structure/stack_queue.md‎

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## Stack 栈
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### [/implement-queue-using-stacks](https://leetcode.cn/problems/implement-queue-using-stacks)
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### [implement-queue-using-stacks](https://leetcode.cn/problems/implement-queue-using-stacks)
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> 请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作(push、pop、peek、empty):
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> 实现 MyQueue 类:
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> 说明:
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> 你 只能 使用标准的栈操作 —— 也就是只有 push to top, peek/pop from top, size, 和 is empty 操作是合法的。
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> 你所使用的语言也许不支持栈。你可以使用 list 或者 deque(双端队列)来模拟一个栈,只要是标准的栈操作即可。
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```python
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class MyQueue:
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def __init__(self):
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self.stack = list()
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def push(self, x: int) -> None:
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self.stack.append(x)
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def pop(self) -> int:
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tmp = None
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if len(self.stack):
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tmp = self.stack[0]
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self.stack = self.stack[1:]
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return tmp
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def peek(self) -> int:
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return self.stack[0]
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def empty(self) -> bool:
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return len(self.stack) == 0
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# Your MyQueue object will be instantiated and called as such:
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# obj = MyQueue()
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# obj.push(x)
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# param_2 = obj.pop()
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# param_3 = obj.peek()
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# param_4 = obj.empty()
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```
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### [/implement-stack-using-queues](https://leetcode.cn/problems/implement-stack-using-queues/)
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> 请你仅使用两个队列实现一个后入先出(LIFO)的栈,并支持普通栈的全部四种操作(push、top、pop 和 empty)。
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```python
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class MyStack:
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def __init__(self):
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self.deq = list()
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def push(self, x: int) -> None:
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self.deq.append(x)
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def pop(self) -> int:
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tmp = None
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if not self.empty():
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tmp = self.deq.pop()
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return tmp
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def top(self) -> int:
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if self.empty():
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return None
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else:
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return self.deq[-1]
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def empty(self) -> bool:
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return len(self.deq) == 0
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# Your MyStack object will be instantiated and called as such:
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# obj = MyStack()
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# obj.push(x)
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# param_2 = obj.pop()
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# param_3 = obj.top()
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# param_4 = obj.empty()
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```
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用collections中的deuqe类
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```python
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from collections import deque
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class MyStack:
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def __init__(self):
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self.deq = deque()
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def push(self, x: int) -> None:
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self.deq.appendleft(x)
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def pop(self) -> int:
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tmp = None
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if len(self.deq):
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tmp = self.deq.popleft()
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return tmp
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def top(self) -> int:
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if self.empty():
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return None
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else:
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return self.deq[0]
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def empty(self) -> bool:
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return len(self.deq) == 0
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# Your MyStack object will be instantiated and called as such:
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# obj = MyStack()
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# obj.push(x)
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# param_2 = obj.pop()
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# param_3 = obj.top()
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# param_4 = obj.empty()
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```
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### [valid-parentheses](https://leetcode.cn/problems/valid-parentheses)
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> 给定一个只包括 '(',')','{','}','[',']' 的字符串 s ,判断字符串是否有效。
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> 有效字符串需满足:
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> 左括号必须用相同类型的右括号闭合。
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> 左括号必须以正确的顺序闭合。
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> 每个右括号都有一个对应的相同类型的左括号。
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括号匹配是栈的拿手好戏。
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```python
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class Solution:
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def paired(self, a, b):
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if (a == '(' and b == ')') or (a == '{' and b == '}' ) or ( a == '[' and b == ']' ):
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return True
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else:
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return False
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def isValid(self, s: str) -> bool:
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s= list(s)
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if len(s) % 2 == 1: return False
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stack = list()
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for i, val in enumerate(s):
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if len(stack) == 0 or not self.paired(stack[-1], val):
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stack.append(val)
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else:
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stack.pop()
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return len(stack) == 0
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```
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### [remove-all-adjacent-duplicates-in-string](https://leetcode.cn/problems/remove-all-adjacent-duplicates-in-string/)
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> 给出由小写字母组成的字符串 S,重复项删除操作会选择两个相邻且相同的字母,并删除它们。
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> 在 S 上反复执行重复项删除操作,直到无法继续删除。
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> 在完成所有重复项删除操作后返回最终的字符串。答案保证唯一。
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```python
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class Solution:
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def removeDuplicates(self, s: str) -> str:
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stack = list()
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for val in s:
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if len(stack) and stack[-1] == val: stack.pop()
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else: stack.append(val)
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return ''.join(stack)
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```
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### [min-stack](https://leetcode-cn.com/problems/min-stack/)
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}
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```
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```python
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class Solution:
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def evalRPN(self, tokens: List[str]) -> int:
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stack = list()
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for val in tokens:
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if len(stack) > 0 and val in ['+', '-', '*', '/']:
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op1 = stack.pop(-1)
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op2 = stack.pop(-1)
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if val == '+': stack.append(op2 + op1)
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elif val == '-': stack.append(op2 - op1)
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elif val == '*': stack.append(op2 * op1)
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else: stack.append(int(op2 / op1))
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else:
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stack.append(int(val))
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return int(stack.pop())
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```
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简洁点
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```python
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class Solution:
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def evalRPN(self, tokens: List[str]) -> int:
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stack = list()
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for val in tokens:
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if len(stack) > 0 and val in ['+', '-', '*', '/']:
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op1 = stack.pop(-1)
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op2 = stack.pop(-1)
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stack.append(int(eval(f'{op2}{val}{op1}')))
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else:
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stack.append(int(val))
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return int(stack.pop())
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```
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### [sliding-window-maximum]()
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### [decode-string](https://leetcode-cn.com/problems/decode-string/)
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> 给定一个经过编码的字符串,返回它解码后的字符串。

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