[leedcode 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)

  class MyStack {
  /*    本题是利用两个队列实现栈，有以下几种方法：
      1、使用两个栈，当pop时，将其中之一队列的0~size()-2 add到备用队列，然后取第一队列的最后值，此种实现有两种方法：
         (1)pop完备用队列可以再复制到第一队列
         (2)增加一个cur标记位，指向当前的有效队列
      2、如果使用一个队列实现，pop时需要计算此时队列中元素个数len，将len-1个元素add到队列的结尾，返回新的队首
      3、使用一个队列，在push元素时，将队列中处在此元素之前的所有值都重新add到队尾，保证对头是最新加入的值，核心代码：
         Q.push(x);
          int size = Q.size() - 1;
          for (int i = 0; i < size; i++) {
              Q.push(Q.front());
              Q.pop();
          }*/
      Map<Integer,LinkedList<Integer>> temp=new HashMap<Integer,LinkedList<Integer>>();
      {
          LinkedList<Integer> t=new LinkedList<Integer>();
          LinkedList<Integer> s=new LinkedList<Integer>();
          temp.put(0,t);
          temp.put(1,s);
      }//注意放到块里，可以初始化
      int cur=0;
      // Push element x onto stack.
      public void push(int x) {
          temp.get(cur).add(x);
      }
  
      // Removes the element on top of the stack.
      public void pop() {
         while(temp.get(cur).size()>1){
             int t=temp.get(cur).remove();
             temp.get(1-cur).add(t);
         }
         temp.get(cur).remove();
         cur=1-cur;
  
      }
  
      // Get the top element.
      public int top() {
           while(temp.get(cur).size()>1){
             int t=temp.get(cur).remove();
             temp.get(1-cur).add(t);
         }
         int res=temp.get(cur).remove();
         temp.get(1-cur).add(res);
         cur=1-cur;
         return res;
      }
  
      // Return whether the stack is empty.
      public boolean empty() {
          return temp.get(cur).isEmpty();
      }
  }