leetcode 232. Implement Queue using Stacks

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～