publicclassMyStack{ Queue<Integer> q;publicMyStack(){// int size():获取队列长度;//boolean add(E)/boolean offer(E):添加元素到队尾;//E remove()/E poll():获取队首元素并从队列中删除;//E element()/E peek():获取队首元素但并不从队列中删除。q =newLinkedList<>(); }publicvoidpush(int...
You must useonlystandard operations of a queue -- which means onlypush to back,peek/pop from front,size, andis emptyoperations are valid. Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as...
LeetCode 225. Implement Stack using Queues 简介:使用队列实现栈的下列操作:push(x) -- 元素 x 入栈;pop() -- 移除栈顶元素;top() -- 获取栈顶元素;empty() -- 返回栈是否为空 Description Implement the following operations of a stack using queues. push(x) -- Push element x onto stack. pop...
int pop()Removes the element on the top of the stack and returns it. int top()Returns the element on the top of the stack. boolean empty()Returnstrueif the stack is empty,falseotherwise. Notes: You must useonlystandard operations of a queue, which means that onlypush to back,peek/pop...
题目链接:https://leetcode.com/problems/implement-stack-using-queues/题目: Implement the following operations of a stack using queues. push(x) -- Push element x onto stack. pop() -- Removes the element on top of the stack. top() -- Get the top element. ...
Implement the following operations of a queue using stacks. push(x) -- Push element x to the back of queue. pop() -- Removes the element from in front of queue. peek() -- Get the front element. empty() -- Return whether the queue is empty. ...
建议和这道题leetcode 232. Implement Queue using Stacks 双栈实现队列 一起学习。 1)取栈顶元素: 返回有元素的队列的首元素 2)判栈空:若队列a和b均为空则栈空 3)入栈:a队列当前有元素,b为空(倒过来也一样)则将需要入栈的元素先放b中,然后将a中的元素依次出列并入列倒b中。(保证有一个队列是空的...
1 审题 LeetCode 225E 栈Stack:后进先出,last-in-first-out LIFO 队列Queue:先进先出,first-in-first-out FIFO 题目要求: 最多使用2个队列,来实现栈; 支持栈的方法: push(x), 把元素 x 推入栈; top/peek(), 返回栈顶元素; pop,移除栈顶元素; ...
self.q_ = Queue() self.top_ = None # @param x, an integer # @return nothing def push(self, x): self.q_.push(x) self.top_ = x # @return nothing def pop(self): for _ in xrange(self.q_.size() - 1): self.top_ = self.q_.pop() self.q_.push(self.to...
queue<int> q2;voidgather(queue<int>& q1, queue<int>& q2){while(!q2.empty()) { q1.push(q2.front()); q2.pop(); } } }; 由于每次push()、top()或者pop()之后都有一个队列为空,因此直接将q1设为有元素的队列即可。可以省去出入队列的时间。