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  • (LeetCode)两个队列来实现一个栈

    原题例如以下:

    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 backpeek/pop from frontsize, 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).
    基本思路是:如果有两个队列Q1和Q2,当二者都为空时。入栈操作能够用入队操作来模拟,能够随便选一个空队列,如果选Q1进行入栈操作。如今如果a,b,c依次入栈了(即依次进入队列Q1)。这时如果想模拟出栈操作,则须要将c出栈。由于在栈顶。这时候能够考虑用空队列Q2,将a,b依次从Q1中出队,而后进入队列Q2,将Q1的最后一个元素c出队就可以。此时Q1变为了空队列。Q2中有两个元素,队头元素为a。队尾元素为b。接下来如果再运行入栈操作,则须要将元素进入到Q1和Q2中的非空队列,即进入Q2队列,出栈的话,就跟前面的一样。将Q2除最后一个元素外所有出队,并依次进入队列Q1,再将Q2的最后一个元素出队就可以。

    Java实现代码例如以下:

    class MyStack {	LinkedList<Integer> queue1 = new LinkedList<Integer>();
    	LinkedList<Integer> queue2 = new LinkedList<Integer>();
        // Push element x onto stack.
        public void push(int x) {
            if(queue1.size()==0&&queue2.size()==0)
            {
            	queue1.offer(x);
            }
            else if(queue1.size()==0)
            {
            	queue2.offer(x);
            }
            else
            {
            	queue1.offer(x);
            }
        }
    
        // Removes the element on top of the stack.
        public void pop() {
        	if(queue1.size()!= 0)
        	{
        		int length = queue1.size();
            for(int i =0;i<length-1;i++)
            {
            	queue2.offer(queue1.poll());
            }
            queue1.poll();
        	}
        	else
        	{
        		int length = queue2.size();
        	for(int i =0;i<length-1;i++)
        	{
        		queue1.offer(queue2.poll());
        	}
        	 queue2.poll();
        	}
        }
        // Get the top element.
        public int top() {
        	if(queue1.size()!= 0)
        	{
        		int length = queue1.size();
            for(int i =0;i<length-1;i++)
            {
            	queue2.offer(queue1.poll());
            }
            int result =  queue1.element();
            queue2.offer(queue1.poll());
    		return result;
        	}
        	else
        	{
        		int length = queue2.size();
        	for(int i =0;i<length-1;i++)
        	{
        		queue1.offer(queue2.poll());
        	}
        	int result = queue2.element();
        	queue1.offer(queue2.poll());
        	return result;
        	}
        }
    
        // Return whether the stack is empty.
        public boolean empty() {
            return queue1.size()==0&&queue2.size()==0;
        }
        
    }


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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5236362.html
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