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LeetCode 232: Implement Queue using Stacks

时间:2017-05-30 20:57:20      阅读:196      评论:0      收藏:0      [点我收藏+]

标签:following   log   queue   pre   cal   rem   note   val   track   

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).

题目要求通过栈来模拟队列的行为。与此类似的还有通过队列模拟栈(http://blog.csdn.net/sunao2002002/article/details/46482673),此题是算法导论第十章的一道题。算法例如以下:

堆栈a和b。a用作入队,b出队

(1)判队满:假设a满且b不为空,则队满

(2)判队空:假设a和b都为空,则队空

(3)入队:首先判队满。

    若队不满:(1)栈a若不满,则直接压入栈a

                        (2)若a满,则将a中的全部元素弹出到栈b中,然后再将元素入栈a

(4)出队:(1)若b空就将a中的全部元素弹出到栈b中。然后出栈

                      (2)b不空就直接从b中弹出元素

代码例如以下:

class Queue {
public:
	// Push element x to the back of queue.
	stack<int> in;
	stack<int> out;
	void push(int x) {
		in.push(x);
	}

	void move(){
		while(!in.empty())
		{
			int x = in.top();
			in.pop();
			out.push(x);
		}
	}
	// Removes the element from in front of queue.
	void pop(void) {
		if (out.empty())
		{
			move();
		}
		if (!out.empty())
		{
			out.pop();
		}
	}

	// Get the front element.
	int peek(void) {
		if (out.empty())
		{
			move();
		}
		if (!out.empty())
		{
			return out.top();
		}

	}

	// Return whether the queue is empty.
	bool empty(void) {
		return in.empty() && out.empty();

	}
};


LeetCode 232: Implement Queue using Stacks

标签:following   log   queue   pre   cal   rem   note   val   track   

原文地址:http://www.cnblogs.com/gccbuaa/p/6921102.html

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