232. Implement Queue using Stacks

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Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

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, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

Constraints:

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, peek, and empty.
• All the calls to pop and peek are valid.

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

class MyQueue {
public:
    stack<int>InStack;
    stack<int>OutStack;

    MyQueue() {

    }
    
    void push(int x) {
        InStack.push(x);
    }
    
    int pop() {
        if(OutStack.empty()){
            while(!InStack.empty()){
                OutStack.push(InStack.top());
                InStack.pop();
            }
        }
        int result=OutStack.top();
        OutStack.pop();
        return result;
    }
    
    int peek() {
        int res=this->pop();
        OutStack.push(res);
        return res;
    }
    
    bool empty() {
        return InStack.empty() && OutStack.empty();
    }
};

注意：

        其实这个没有什么算法难点，主要以下两点：

        1.使用两个stack，一个in一个out

        2.注意函数是void还是int，即是否需要返回值

本文标签： implementStacksQueue