LeetCode-232. Implement Queue using Stacks [C++][Java]

LeetCode-232. Implement Queue using Stackshttps://leetcode.com/problems/implement-queue-using-stacks/

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.

【C++】

class MyQueue {
    stack in, out;
public:
    MyQueue() {}
    
    void push(int x) {
        in.push(x);
    }
    
    int pop() {
        in2out();
        int x = out.top();
        out.pop();
        return x;
    }
    
    int peek() {
        in2out();
        return out.top();
    }
    
    void in2out() {
        if (out.empty()) {
            while (!in.empty()) {
                int x = in.top();
                in.pop();
                out.push(x);
            }
        }
    }
    
    bool empty() {
        return in.empty() && out.empty();
    }
};

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue* obj = new MyQueue();
 * obj->push(x);
 * int param_2 = obj->pop();
 * int param_3 = obj->peek();
 * bool param_4 = obj->empty();
 */

【Java】

class MyQueue {
    private Stack in = new Stack<>();
    private Stack out = new Stack<>();

    public MyQueue() {}
    
    public void push(int x) {
        in.push(x);
    }
    
    public int pop() {
        in2out();
        int x = out.pop();
        return x;
    }
    
    public int peek() {
        in2out();
        return out.peek();
    }
    
    void in2out() {
        if (out.empty()) {
            while (!in.empty()) {
                int x = in.pop();
                out.push(x);
            }
        }
    }
    
    public boolean empty() {
        return in.empty() && out.empty();
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

参考文献

【1】Java的Stack类_Java硬件工程师的博客-CSDN博客_java stack类