leetcode 232. Implement Queue using Stacks

最新推荐文章于 2021-03-25 10:39:22 发布

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本文介绍了一种使用栈来实现队列的方法。通过两个栈分别用于入队和出队操作，实现了队列的基本功能，包括push、pop、peek和empty等操作。文章还提供了完整的代码示例，并附带了LeetCode上的题解。

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.

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).

解题思路：

本题比较简单，主要考察对队列和栈的理解，然后注意摆脱只能使用一个栈的思维定势就可以了

class Queue {
public:
    // Push element x to the back of queue.
    void push(int x) {
        in.push(x);
    }

    // Removes the element from in front of queue.
    void pop(void) {
        if(out.empty())
            trans();
        out.pop();
    }

    // Get the front element.
    int peek(void) {
        if(out.empty())
            trans();
        return out.top();
    }

    // Return whether the queue is empty.
    bool empty(void) {
        return out.empty() && in.empty();
    }
private:
    stack<int> in, out;
    void trans(){
        while(!in.empty()){
            out.push(in.top());
            in.pop();
        }
    }
};

另附上 leetcode上的题解，上面有对时空复杂度进行分析，时空复杂度在本题不是很直观(这涉及到均摊复杂度的概念 Amortized Analysis )，可以看一下～