[LeetCode] 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).

     

     这道题就不说了，和之前说过的那道implement Queue using Stack的思路差不多。

     还是建立两个Queue之间互相转换就好，记得所写的算法运行后必须保证两个queue的其中一个队列为空。

     我的答案这里把push写成了offer然后pop写成了poll，不过leetcode还是识别了哈哈。

     因为我写的时候有参考java platform standard（http://docs.oracle.com/javase/7/docs/api/java/util/Queue.html#peek()）这里面还是写的offer/poll所以就……。

class MyStack {
    LinkedList<Integer> a=new LinkedList<Integer>();
    LinkedList<Integer> b=new LinkedList<Integer>();
    // Push element x onto stack.
    public void push(int x) {
        if(a.size()==0){
            a.offer(x);
        }else{
           if(a.size()>0){
               b.offer(x);
               int size=a.size();
               while(size>0){
                   b.offer(a.poll());
                   size--;
               }
           }else if(b.size()>0){
               a.offer(x);
               int size=b.size();
               while(size>0){
                   a.offer(b.poll());
                   size--;
               }
           }
            
        }
    }

    // Removes the element on top of the stack.
     public void pop() {
         if(a.size()>0){
            a.poll();
         }else if(b.size()>0){
             b.poll();
         }
        
    }

    // Get the top element.i
    public int top() {
        if(a.size()>0){
            return a.peek();
        }else if(b.size()>0){
            return b.peek();
        }
        return 0;
    }

    // Return whether the stack is empty.
    public boolean empty() {
        return a.isEmpty()&b.isEmpty();
    }
}