Question:
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.
Notes:
- You must use only standard operations of a stack -- which means only
push to top
,peek/pop from top
,size
, andis empty
operations are valid. - Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
Analysis:
问题描述:用队列模仿一个栈。
思路:用两个队列模仿一个栈。每次要pop或者peek时,使用队列倒换一下,剩下最后一个元素单独处理。当且仅当两个队列都为空时,栈才为空。
Answer:
class MyStack { Queue<Integer> q1 = new LinkedList<Integer>(); Queue<Integer> q2 = new LinkedList<Integer>(); // Push element x onto stack. public void push(int x) { q1.offer(x); } // Removes the element on top of the stack. public void pop() { if(!q1.isEmpty()) { while(q1.size() > 1) { int i = q1.poll(); q2.offer(i); } q1.poll(); } else { while(q2.size() > 1) { int i = q2.poll(); q1.offer(i); } q2.poll(); } } // Get the top element. public int top() { if(!q1.isEmpty()) { while(q1.size() > 1) { int i = q1.poll(); q2.offer(i); } int i = q1.poll(); q2.offer(i); return i; } else { while(q2.size() > 1) { int i = q2.poll(); q1.offer(i); } int i = q2.poll(); q1.offer(i); return i; } } // Return whether the stack is empty. public boolean empty() { if(q1.size() == 0 && q2.size() == 0) return true; return false; } }
时间: 2024-10-13 23:25:47