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[Java教程][LeetCode] Implement Queue using Stacks


  国际惯例,先上题目:

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 toppeek/pop from topsize, 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).

  

     这道题个人觉得属于比较有意思的一道题><还有一道类似的用queue来完成stack的,不过我还没有做到,暂且不说。   

     这道题比较好想到的方法是双stack法,因为刚好stack和queue的顺序是相反的。

     试想建立两个stack,当把所有数据从stack a中转移到b的时候,其数据排列就会发生变化,这样一来就很好理解了。

     不过也可以只建立一个stack,这里就需要swap来辅助,这个会比较节省时间一点。

     不过我暂时只写了双stack的用法,下次有机会写下单stack的哈哈。

     个人觉得这道题难点就在第一个method上,后面都很简单,直接套用stack的method即可。

class MyQueue {    Stack<Integer> a=new Stack<Integer>();  Stack<Integer> b=new Stack<Integer>();    // Push element x to the back of queue.  public void push(int x) {    //specail case    if(a.isEmpty()){      a.push(x);    }else{      while(!a.isEmpty()){        b.push(a.pop());      }      a.push(x);      while(!b.isEmpty()){        a.push(b.pop());      }    }      }  // Removes the element from in front of queue.  public void pop() {    a.pop();  }  // Get the front element.  public int peek() {    return a.peek();  }  // Return whether the queue is empty.  public boolean empty() {    return a.isEmpty();  }}