LeetCode #232 - Implement Queue using Stacks

最新推荐文章于 2024-10-30 00:15:00 发布

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最新推荐文章于 2024-10-30 00:15:00 发布

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本文介绍了一种使用两个栈来实现队列的方法，包括push、pop、peek和empty操作。通过将元素推入第一个栈，然后在需要时将元素转移到第二个栈以实现队列的先进先出特性。

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题目描述：

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.

Example:

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).

class MyQueue {
public:
    /** Initialize your data structure here. */
    MyQueue() {}
    
    /** Push element x to the back of queue. */
    void push(int x) {
        s1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        int result=0;
        if(!s2.empty())
        {
            result=s2.top();
            s2.pop();
            return result;
        }
        else
        {
            while(!s1.empty())
            {
                s2.push(s1.top());
                s1.pop();
            }
            result=s2.top();
            s2.pop();
            return result;
        }
    }
    
    /** Get the front element. */
    int peek() {
        int result=0;
        if(!s2.empty())
        {
            result=s2.top();
            return result;
        }
        else
        {
            while(!s1.empty())
            {
                s2.push(s1.top());
                s1.pop();
            }
            result=s2.top();
            return result;
        }
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return s1.empty()&&s2.empty();
    }

private:
    stack<int> s1; // push进来的元素先进入s1
    stack<int> s2; // 需要pop时，从s2中获取，当s2为空时，从s1获取元素
};