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LeetCode Implement Queue using Stacks

时间:2015-07-21 17:22:00      阅读:109      评论:0      收藏:0      [点我收藏+]

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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).
思路分析:这题和几年前写过的一篇面经博文是相同的题目: 面试题研究 用两个栈模拟实现队列。基本做法很简单,用两个栈就可以模拟一个队列,基本思路是两次后进先出 = 先进先出,元素入队列总是入A栈,元素出队列如果B栈不为空直接弹出B栈头元素;如果B栈为空就把A栈元素出栈全部压入B栈,再弹出B栈头,这样就模拟出了一个队列。核心就是保证每个元素出栈时都经过了A,B两个栈,这样就实现了两次后进先出=先进先出。

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

    // Removes the element from in front of queue.
    public void pop() {
        if(!stackB.isEmpty()) {
            stackB.pop();
        } else {
            while(!stackA.isEmpty()){
                stackB.push(stackA.pop());
            }
            stackB.pop();
        }
    }

    // Get the front element.
    public int peek() {
        if(!stackB.isEmpty()) {
            return stackB.peek();
        } else {
            while(!stackA.isEmpty()){
                stackB.push(stackA.pop());
            }
            return stackB.peek();
        }
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return (stackA.isEmpty() && stackB.isEmpty());
    }
}


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LeetCode Implement Queue using Stacks

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原文地址:http://blog.csdn.net/yangliuy/article/details/46985947

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