Lettcode_252_Implement Stack using Queues

本文是在学习中的总结，欢迎转载但请注明出处：http://blog.csdn.net/pistolove/article/details/48598773

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.
• empty() -- Return whether the stack is empty.

Notes:

• You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations 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 you use only standard operations of a queue.
• You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

思路：

（1）题意为运用队列来实现栈，主要包括push()、pop()、top()、empty()方法的实现。

（2）该题的思路和“运用栈来实现队列”类似。首先还是得了解栈和队列的特性，这里再复习一遍；栈：先进后出，只能在栈顶添加和删除元素，队列：先进先出，只能在队尾添加元素，从队头移除元素。想用队列实现栈，主要需考虑到如何将先放入的元素先出栈；这里需要用两个队列来实现，其中一个队列在执行操作后为空，另一个队列存放元素。当有元素加入时，如果某一队列不为空，则将元素加入该队列中；当有元素出栈时，此时需从不为空的队列list中将除去队尾元素的其余元素全部加入另一个空的队列list0中，这样，list队列中的队尾元素(对应栈顶元素)就没有被加入到list0中，然后将list置为一个空的队列，即完成出栈操作。其余的操作比较简单，请参见代码。

（3）详情见下方代码。希望本文对你有所帮助。

算法代码实现如下：

package leetcode;

import java.util.LinkedList;
import java.util.List;

/**
 * 
 * @author liqqc
 *
 */
public class Implement_Stack_using_Queues {

	// Push element x onto stack.

	private List<Integer> list = new LinkedList<Integer>();
	private List<Integer> list0 = new LinkedList<Integer>();

	public void push(int x) {
		if (list.size() == 0 && list0.size() == 0) {
			list.add(x);
			return;
		}

		if (list.size() != 0) {
			list.add(x);
			return;
		}

		if (list0.size() != 0) {
			list0.add(x);
		}

	}

	// Removes the element on top of the stack.
	public void pop() {

		if (list.size() != 0 && list0.size() == 0) {
			for (int i = 0; i < list.size() - 1; i++) {
				list0.add(list.get(i));
			}
			list.clear();
			return;
		}

		if (list.size() == 0 && list0.size() != 0) {
			for (int i = 0; i < list0.size() - 1; i++) {
				list.add(list0.get(i));
			}
			list0.clear();
		}

	}

	// Get the top element.
	public int top() {
		if (list.size() != 0 && list0.size() == 0) {
			return list.get(list.size() - 1);
		}

		if (list.size() == 0 && list0.size() != 0) {
			return list0.get(list0.size() - 1);
		}

		return -1;
	}

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

		return list0.size() == 0 && list.size() == 0;
	}
}