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.

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.

Example 1:

<pre style="box-sizing: border-box; font-family: SFMono-Regular, Consolas, "Liberation Mono", Menlo, Courier, monospace; font-size: 13px; margin-top: 0px; margin-bottom: 1em; overflow: auto; background: var(--dark-bg); border-color: rgb(51, 51, 51); color: var(--font) !important; padding: 10px 15px; line-height: 1.6; border-radius: 3px; white-space: pre-wrap;">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
</pre>

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.

我的答案：stack的基本操作

class MyQueue {
public:
    /** Initialize your data structure here. */
    stack<int> s;
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        s.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        stack<int> s_;
        while (s.size() > 1) {
            s_.push(s.top());
            s.pop();
        }
        int ans = s.top();
        s.pop();
        
        while (s_.size() > 0) {
            s.push(s_.top());
            s_.pop();
        }    
            
        return ans;
    }
    
    /** Get the front element. */
    int peek() {
        stack<int> s_;
        while (s.size() > 1) {
            s_.push(s.top());
            s.pop();
        }
        int ans = s.top();
        
        while (s_.size() > 0) {
            s.push(s_.top());
            s_.pop();
        }    
            
        return ans;
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return s.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();
 */

0ms >100% submission