/*how to use two stacks to implement the queue: offer, poll, peek,size, isEmpty offer(3) offer(2) poll() offer(1) peek() offer(6) poll() poll() 3 2 2 1 in (3 2) (1 6) out (2) (3) 6 (1) stack1(in): is the only stack to store new elements when adding a new ...
用栈实现队列 正如标题所述,你需要使用两个栈来实现队列的一些操作。 队列应支持push(element),pop() 和 top(),其中pop是弹出队列中的第一个(最前面的)元素。 pop和top方法都应该返回第一个元素的值。 样例 比如 push(1), pop(), push(2), push(3), top(), pop(),你应该返回1,2和2 挑战 仅...
class QueueTwoStacks(object): # Implement the enqueue and dequeue methods def enqueue(self, item): pass def dequeue(self): pass # Tests class Test(unittest.TestCase): def test_basic_queue_operations(self): queue = QueueTwoStacks() queue.enqueue(1) queue.enqueue(2) queue.enqueue(3) actual...
http://www.lintcode.com/en/problem/implement-queue-by-two-stacks/ 【题目解析】 用两个Stack来实现一个Queue,可以考虑到push()时,几乎与Queue中的offer()一样,都是加在末尾,区别是当Stack pop()时,取出的是最近加入(newest)的元素,而Queue用poll()则是将最老(oldest)的元素取出。使用2个Stack,可以将...
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1. 232 Implement Queue using Stacks 1.1 问题描写叙述 使用栈模拟实现队列。模拟实现例如以下操作: push(x). 将元素x放入队尾。 pop(). 移除队首元素。 peek(). 获取队首元素。 empty(). 推断队列是否为空。 注意:仅仅能使用栈的标准操作,push,pop,size和empty函数。
Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push,peek,pop, andempty). Implement theMyQueueclass: void push(int x)Pushes element x to the back of the queue. ...
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. ...
Implement Queue by Two Stacks As the title described, you should only use two stacks to implement a queue's actions. The queue should supportpush(element),pop()andtop()where pop is pop the first(a.k.a front) element in the queue. ...
A simple solution would be to divide the array into two halves and allocate each half to implement two stacks. In other words, for an arrayAof sizen, the solution would allocateA[0, n/2]memory for the first stack andA[n/2+1, n-1]memory for the second stack. The problem with this...