#40 Implement Queue by Two Stacks

最新推荐文章于 2023-07-29 01:49:56 发布

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#lintcode #stack #queue

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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和top操作，并确保平均时间复杂度为O(1)。示例展示了如何通过两个栈进行元素的压入和弹出操作。

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题目描述：

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.

Mycode（AC = 37ms）：

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

    Queue() {
        // do intialization if necessary
    }

    void push(int element) {
        // write your code here
        stack1.push(element);
    }
    
    int pop() {
        // write your code here
        while (!stack1.empty()) {
            stack2.push(stack1.top());
            stack1.pop();
        }
        
        int val = stack2.top();
        stack2.pop();
        while(!stack2.empty()) {
            stack1.push(stack2.top());
            stack2.pop();
        }
        
        return val;
    }

    int top() {
        // write your code here
        while (!stack1.empty()) {
            stack2.push(stack1.top());
            stack1.pop();
        }
        
        int val = stack2.top();
        while(!stack2.empty()) {
            stack1.push(stack2.top());
            stack2.pop();
        }
        
        return val;
    }
};