peek() 需要考虑; 如果stack 2 非空,直接返回其栈顶元素; 如果stack 2 为空,则把 stack 1的元素全部移动到 stack 2,并返回栈顶元素 pop() 类似于peek,但是在返回的时候是移除操作并返回该元素。 1 读题 LeetCode 232E 用栈实现队列 Implement Queue using Stacks 2 Python 解题 正如咱们前面那个问题,list...
x:int)->None:## Push x onto stackself.l1.append(x)## The right of the list = the top of the stack = the back of the queuedefpop(self)->int:## pop 抽取## Removes the top element and returnifself.l1:whilelen(self.l1)>1:self.l2.append(...
outStack []int// pop, 作为对外出栈 } // initialize data struct func Constructor() MyQueue { returnMyQueue{ inStack: make([]int,0),// push outStack: make([]int,0),// pop } } // push,直接压入instack func (this*MyQueue) Push(x int) { this.inStack = append(this.inStack, x...
return True if (len(self.inStack)+len(self.outStack))==0 else False
Python解法 下面的python代码是把stack2当做是和队列顺序一样的,这样的话,如果stack2不空,那么久弹出元素就行。否则,如果stack1中有元素,那么在做push和pop的时候,需要先把stack1中的元素颠倒到stack2中。 AI检测代码解析 class MyQueue(object): def __init__(self): ...
class Stack: # initialize your data structure here. def __init__(self): self.q_ = Queue() # @param x, an integer # @return nothing def push(self, x): self.q_.push(x) for _ in xrange(self.q_.size() - 1): self.q_.push(self.q_.pop()) # @return nothing ...
Okay, if you’re threading, you can’t uselistfor a stack and you probably don’t want to usedequefor a stack, so howcanyou build a Python stack for a threaded program? The answer is in thequeuemodule,queue.LifoQueue. Remember how you learned that stacks operate on the Last-In/First...
Dequeue an element from the queue. C++ Java Python #include<bits/stdc++.h> using namespace std; class Stack { queue<int>q; public: void push(int val); void pop(); int top(); bool empty(); }; void Stack::push(int val) { int s = q.size(); q.push(val); for (int i=0...
使用以下文章中的内容时候,如果遇见了 The client is using an unsupported version of the Socket.IO or protocols Error 这个错误,那就是引用的版本问题。参考文档:https://stackoverflow.com/questions/66069215/the-client-is-using-an-unsupported-version-of-the-socket-io-or-engine-io-protoco...
Linux doesn't offer an API for determining the current stack bounds, at least not that I am aware of. This means we will need to probe the stack using a SIGSEGV handler and longjmp/setjump. It is somewhat ugly but should work. Actually, ...