C#LeetCode刷题之#232-用栈实现队列​​​​​​​​​​​​​​(Implement Queue using Stacks)

问题

使用栈实现队列的下列操作:

push(x) — 将一个元素放入队列的尾部。
pop() — 从队列首部移除元素。
peek() — 返回队列首部的元素。
empty() — 返回队列是否为空。

MyQueue queue = new MyQueue();

queue.push(1);

queue.push(2);  

queue.peek();  // 返回 1

queue.pop();   // 返回 1

queue.empty(); // 返回 false

说明:

你只能使用标准的栈操作 — 也就是只有 push to top, peek/pop from top, size, 和 is empty 操作是合法的。
你所使用的语言也许不支持栈。你可以使用 list 或者 deque(双端队列)来模拟一个栈,只要是标准的栈操作即可。
假设所有操作都是有效的 (例如,一个空的队列不会调用 pop 或者 peek 操作)。

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.

MyQueue queue = new MyQueue();

queue.push(1);

queue.push(2);  

queue.peek();  // returns 1

queue.pop();   // returns 1

queue.empty(); // returns false

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).

示例

public class Program {

    public static void Main(string[] args) {
        var queue = new MyQueue();

        queue.Push(1);
        queue.Push(2);
        queue.Push(3);

        Console.WriteLine(queue.Peek());
        Console.WriteLine(queue.Pop());
        Console.WriteLine(queue.Empty());

        Console.ReadKey();
    }

    public class MyQueue {

        private Stack<int> _stack = null;

        public MyQueue() {
            _stack = new Stack<int>();
        }

        public void Push(int x) {
            //基本思路是反转原栈
            var stack = new Stack<int>();
            stack.Push(x);
            var reverse = _stack.Reverse().ToList();
            foreach(var elemet in reverse) {
                stack.Push(elemet);
            }
            _stack = stack;
        }

        public int Pop() {
            return _stack.Pop();
        }

        public int Peek() {
            return _stack.Peek();
        }

        public bool Empty() {
            return _stack.Count == 0;
        }

    }

}

以上给出1种算法实现,以下是这个案例的输出结果:

1
1
False

分析:

显而易见,因为部分运行库的使用,Push 的时间复杂度应当为: O(n)​ ,其它方法的时间复杂度应当为: O(1) 。

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