LeetCode——Implement Queue using Stacks

Description:

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 top, peek/pop from top, size, 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).

用栈来实现队列的功能。类似的还有用队列实现栈的功能。

思路:栈和队列对数据的处理是不同的。栈是先进后出(FILO)队列是先进先出(FIFO)。所以要用两个栈来维护一个队列。一个数据栈,一个暂存数据栈。数据栈用来存储数据,暂存数据栈用来把数据栈中的数据首尾交换,模拟队列的数据操作。

 

代码:

class MyQueue {
    Stack<Integer> stack1;
    Stack<Integer> stack2;
    
    public MyQueue() {
        stack1 = new Stack();
        stack2 = new Stack();
    }
    
    // Push element x to the back of queue.
    public void push(int x) {
        stack1.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        //把栈中的元素移到另一个栈中,首尾倒序。
        while(!stack1.empty()) {
            stack2.push(stack1.pop());
        }
        //移除堆首元素。
        stack2.pop();
        //还原数据队列。
        while(!stack2.empty()) {
            stack1.push(stack2.pop());
        }
    }

    // Get the front element.
    public int peek() {
        //把栈中的元素移到另一个栈中,首尾倒序。
        while(!stack1.empty()) {
            stack2.push(stack1.pop());
        }
        //获取堆首元素。
        int front = stack2.peek();
        //还原数据队列。
        while(!stack2.empty()) {
            stack1.push(stack2.pop());
        }
        return front;
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return stack1.empty();
    }
}

 

posted @ 2015-07-30 09:35  Pickle  阅读(202)  评论(0编辑  收藏  举报