代码随想录算法训练营第十天 | 232. 用栈实现队列、225. 用队列实现栈、20. 有效的括号、1047. 删除字符串中的所有相邻重复项

232. 用栈实现队列

双栈实现队列，一个管进一个管出即可。

class MyQueue:

    def __init__(self):
        self.stackIn = []
        self.stackOut = []

    def push(self, x: int) -> None:
        self.stackIn.append(x)

    def pop(self) -> int:
        if self.empty():
            return None
        
        if self.stackOut:
            return self.stackOut.pop()
        else:
            while self.stackIn:
                self.stackOut.append(self.stackIn.pop())
            return self.stackOut.pop()

    def peek(self) -> int:
        if self.empty():
            return None
        
        res = self.pop()
        self.stackOut.append(res)
        return res
        
    def empty(self) -> bool:
        return not self.stackIn and not self.stackOut


# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()

225. 用队列实现栈

一个队列就可以实现栈

from collections import deque

class MyStack:

    def __init__(self):
        self.que = deque()

    def push(self, x: int) -> None:
        self.que.append(x)       

    def pop(self) -> int:
        if self.empty():
            return None
        for i in range(len(self.que) - 1):
            self.que.append(self.que.popleft())
        return self.que.popleft()

    def top(self) -> int:
        if self.empty():
            return None
        return self.que[-1]

    def empty(self) -> bool:
        return not self.que

20. 有效的括号

经典栈应用，判断括号闭合。
但是照着测试数据才能考虑完全…

class Solution:
    def isValid(self, s: str) -> bool:
        h = {'(': ')', '[': ']', '{': '}'}
        stack = []
        for char in s:
            if char in h:
                stack.append(h[char])
            else:            
                if not stack:
                    return False
                
                elif char == stack[-1]:
                    stack.pop()
                
                else:
                    return False
            
        return not stack

合并一下条件：

class Solution:
    def isValid(self, s: str) -> bool:
        h = {'(': ')', '[': ']', '{': '}'}
        stack = []
        for char in s:
            if char in h:
                stack.append(h[char])
            else:            
                if not stack or char != stack[-1]:
                    return False
                else:
                    stack.pop()
            
        return not stack

1047. 删除字符串中的所有相邻重复项

每次循环一个字符，查看栈顶，如果栈非空且栈顶与当前遍历字符相同，弹出栈顶且不加入新元素；否则加入新元素。

class Solution:
    def removeDuplicates(self, s: str) -> str:
        stack = []
        for char in s:
            if stack and char == stack[-1]:
                stack.pop()
            else:
                stack.append(char)
        return "".join(stack)