两个队列实现一个栈

问题描述：

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).

link: https://leetcode.com/problems/implement-stack-using-queues/

解决思路：

<思路1>

假设有两个队列，分别为q1和q2. 对数据元素进行操作时使用q1，将q2作为辅助的数据缓存的队列。具体做法如下：

(1) push时，直接将需要push的元素加到q1中；

(2) pop时，首先检查q1是否为空，如果不为空则将q1中的元素都poll到q2中，直到q1中的元素剩余一个为止，剩下的这个元素即为需要pop的元素，poll即可；

(3) peek时，与(2)中的pop类似，区别在于最后不poll，保留该值，最终返回即可。

(4) isEmpty()只需要检查q1.size()是否为0；

 

<代码>

class MyStack {
    private Queue<Integer> q1;
    private Queue<Integer> q2;
    
    public MyStack() {
        q1 = new LinkedList<Integer>();
        q2 = new LinkedList<Integer>();
    }
    
    // Push element x onto stack.
    public void push(int x) {
        q1.offer(x);
    }

    // Removes the element on top of the stack.
    public void pop() {
        if (q1.size() == 0) {
            return ;
        }
        while (q1.size() > 1) {
            q2.offer(q1.poll());
        }
        q1.poll();
        while (q2.size() > 0) {
            q1.offer(q2.poll());
        }
    }

    // Get the top element.
    public int top() {
        if (q1.size() == 0) {
            return -1;
        }
        while (q1.size() > 1) {
            q2.offer(q1.poll());
        }
        int res = q1.poll();
        q2.offer(res);
        while (q2.size() > 0) {
            q1.offer(q2.poll());
        }
        return res;
    }

    // Return whether the stack is empty.
    public boolean empty() {
        return q1.size() == 0;
    }
}

 

<思路2>

思路1中每次pop操作之后，都要将q2中的元素倒回q1中，这样做的的优点可以使得push操作更简单，但是缺点在于效率不高。下面考虑另一种思路，使用两个指针pushTmp和tmp来标记进出栈操作对应的队列和中间缓存的队列，只要这两种类型的队列确定了，对应的pop和push操作和思路1中的是一样的。tmp对应的队列通常为空。

<代码>

class MyStack {
    private Queue<Integer> q1;
    private Queue<Integer> q2;
    
    private Queue<Integer> pushTmp;
    private Queue<Integer> tmp;
    
    public MyStack() {
        q1 = new LinkedList<Integer>();
        q2 = new LinkedList<Integer>();
    }
    
    // Push element x onto stack.
    public void push(int x) {
        if (q1.size() == 0) {
            pushTmp = q2;
        } else {
            pushTmp = q1;
        }
        pushTmp.offer(x);
    }

    // Removes the element on top of the stack.
    public void pop() {
        if (q1.size() == 0) {
            tmp = q1;
            pushTmp = q2;
        } else {
            tmp = q2;
            pushTmp = q1;           
        }
        if (pushTmp.size() == 0) {
            return ;
        }
        while (pushTmp.size() > 1) {
            tmp.offer(pushTmp.poll());
        }
        pushTmp.poll();
    }

    // Get the top element.
    public int top() {
        if (q1.size() == 0) {
            tmp = q1;
            pushTmp = q2;
        } else {
            tmp = q2;
            pushTmp = q1;           
        }
        if (pushTmp.size() == 0) {
            return -1;
        }
        while (pushTmp.size() > 1) {
            tmp.offer(pushTmp.poll());
        }
        int res = pushTmp.poll();
        tmp.offer(res);
        return res;
    }

    // Return whether the stack is empty.
    public boolean empty() {
        return q1.size() == 0 && q2.size() == 0;
    }
}

 

拓展问题：两个栈实现一个队列

<思路1>

假设有两个栈分别为S1和S2，按照以下方法来模拟队列：

(1) 元素进队列：直接将元素push到S1中；

(2) 元素出队列：首先检查S2是否为空，如果不为空则S2.pop()即可；如果为空，则将S1中的所有元素都pop并且push到S2中(如果S1为空则抛出抛出异常)。

 

<代码>

class MyQueue{
	private Stack<Integer> s1;
	private Stack<Integer> s2;
	
	public MyQueue() {
		s1 = new Stack<Integer>();
		s2 = new Stack<Integer>();
	}
	
	public void offer(int elem){
		s1.push(elem);
	}
	
	public int poll() {
		if (!s2.isEmpty()) {
			return s2.pop();
		}
		if (s1.isEmpty()) {
			return -1;
		}
		while (!s1.isEmpty()) {
			s2.push(s1.pop());
		}
		return s2.pop();
	}
	
	public int peek() {
		if (!s2.isEmpty()) {
			return s2.peek();
		}
		while (!s1.isEmpty()) {
			s2.push(s1.pop());
		}
		return s2.peek();
	}
}