LeetCode 225: Implement Stack using Queues

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).

用两个队列模拟一个堆栈:

队列a和b

(1)取栈顶元素: 返回有元素的队列的首元素

(2)判栈空:若队列a和b均为空则栈空

(3)入栈:a队列当前有元素,b为空(倒过来也一样)则将需要入栈的元素先放b中,然后将a中的元素依次出列并入列倒b中。(保证有一个队列是空的)

(4)出栈:将有元素的队列出列即可。

比如先将1插入队a中 ,现在要将2入栈,则将2插入b总然后将a中的1出列入到b中,b中的元素变为 2 ,1

a为空,现在要压入3 则将3插入a中 ,依次将b中的2 ,1 出列并加入倒a中 ,a中的元素变为 3,2,1 b为空

算法保证在任何时候都有一队列为空

代码如下:

class Stack {
public:
	// Push element x onto stack.
	queue queue1;
	queue queue2;
	void push(int x) {
		if (queue1.empty())
		{
			queue1.push(x);
			while(!queue2.empty()){
				int tmp = queue2.front();
				queue2.pop();
				queue1.push(tmp);
			}
		}else{
			queue2.push(x);
			while(!queue1.empty()){
				int tmp = queue1.front();
				queue1.pop();
				queue2.push(tmp);
			}
		}
	}

	// Removes the element on top of the stack.
	void pop() {
		if (!queue1.empty())
			queue1.pop();
		if (!queue2.empty())
			queue2.pop();
	}

	// Get the top element.
	int top() {
		if (!queue1.empty())
			return queue1.front();
		if (!queue2.empty())
			return queue2.front();
	}

	// Return whether the stack is empty.
	bool empty() {
		return queue1.empty() && queue2.empty();
	}
};


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