LeetCode 232 Implement Queue using Stacks

题目描述

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, and is 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).

分析

参考:LeetCode 225 Implement Stack using Queues

非常经典的题目,定义两个栈模拟队列。

  1. push向stack1
  2. pop从stack2,如果stack2为空,先将stack1的元素放入stack2

代码

    class MyQueue {

        Stack<Integer> stack1 = new Stack<Integer>();
        Stack<Integer> stack2 = new Stack<Integer>();

        // Push element x to the back of queue.
        public void push(int x) {
            stack1.push(x);
        }

        // Removes the element from in front of queue.
        public void pop() {
            if (stack2.isEmpty()) {
                if (stack1.isEmpty()) {
                    throw new IllegalStateException();
                }
                while (!stack1.isEmpty()) {
                    stack2.push(stack1.pop());
                }
            }
            stack2.pop();
        }

        // Get the front element.
        public int peek() {
            if (stack2.isEmpty()) {
                if (stack1.isEmpty()) {
                    throw new IllegalStateException();
                }
                while (!stack1.isEmpty()) {
                    stack2.push(stack1.pop());
                }
            }
            return stack2.peek();
        }

        // Return whether the queue is empty.
        public boolean empty() {

            if (stack1.isEmpty() && stack2.isEmpty()) {
                return true;
            }

            return false;
        }
    }
    原文作者:_我们的存在
    原文地址: https://blog.csdn.net/Yano_nankai/article/details/50260479
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