[LeetCode]Implement Queue using Stacks

最新推荐文章于 2021-11-08 15:11:52 发布

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本文介绍了一种使用两个栈来模拟队列行为的方法。通过将元素压入一个栈并在需要时将其依次移动到另一个栈来实现队列的先进先出特性。这种方式仅使用标准的栈操作，包括压栈、弹栈、获取栈顶元素等。

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).
［思路］利用两个堆栈实现队列结构。最容易想到的，push时，push入一个堆栈中。pop时，把堆栈内所有元素弹出到另一个堆栈中（利用两次后入先出实现先入先出），再把元素弹出。特别注意一点，当第二个堆栈不是全空的时候，pop出堆栈顶即可；全空的时候要把第一个堆栈中元素全部弹出到另一个堆栈。不必把第二个堆栈中元素全部弹回第一个堆栈。

class Queue {
private:
    stack<int> stack1;
    stack<int> stack2;
public:
    // Push element x to the back of queue.
    void push(int x) {
        stack1.push(x);
    }

    // Removes the element from in front of queue.
    void pop(void) {
        if(!stack2.empty()){
            stack2.pop();
        }
        else{
            while(!stack1.empty()){
                int temp = stack1.top();
                stack1.pop();
                stack2.push(temp);
            }
            stack2.pop();
        }
        
    }

    // Get the front element.
    int peek(void) {
         if(!stack2.empty()){
            return stack2.top();
        }
        else{
            while(!stack1.empty()){
                int temp = stack1.top();
                stack1.pop();
                stack2.push(temp);
            }
            return stack2.top();
        }
    }

    // Return whether the queue is empty.
    bool empty(void) {
        return stack1.empty()&&stack2.empty();
    }
};