lintcode: Implement Queue by Two Stacks

最新推荐文章于 2021-05-20 01:29:12 发布

martin_liang

最新推荐文章于 2021-05-20 01:29:12 发布

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分类专栏：算法 C++/C leetcode

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本文介绍了一种使用两个栈来实现队列的方法，并确保了push、pop和top操作的平均时间复杂度为O(1)。通过巧妙地在两个栈间转移元素，实现了队列的基本功能。

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As the title described, you should only use two stacks to implement a queue's actions.

The queue should support push(element), pop() and top() where pop is pop the first(a.k.a front) element in the queue.

Both pop and top methods should return the value of first element.

Have you met this question in a real interview?

Yes

Example

push(1)
pop()     // return 1
push(2)
push(3)
top()     // return 2
pop()     // return 2

Challenge

implement it by two stacks, do not use any other data structure and push, pop and top should be O(1) by AVERAGE.

class Queue {
public:
    stack<int> stack1;
    stack<int> stack2;

    Queue() {
        // do intialization if necessary
    }

    void push(int element) {
        // write your code here
        while (stack2.size() > 0)
        {
            stack1.push(stack2.top());
            stack2.pop();
        }
        
        stack1.push(element);
    }
    
    int pop() {
        
        while (stack1.size() > 0)
        {
            stack2.push(stack1.top());
            stack1.pop();
        }
        
        if (stack2.size() > 0)
        {
            int tmp = stack2.top();
            stack2.pop();
            return tmp;
        }
        
        return -1;
        // write your code here
    }

    int top() {
        
        while (stack1.size() > 0)
        {
            stack2.push(stack1.top());
            stack1.pop();
        }
        
        if (stack2.size() > 0)
        {
            return stack2.top();
        }
        
        return -1;
        // write your code here
    }
};