Leetcode 232 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.

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:

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.

1 栈模拟1

设立两个栈stk1和stk2, stk1储存倒置顺序的元素, stk2为空.

push的时候先把stk1所有元素转移到stk2, 然后把x放到stk1栈底, 然后再把stk2剩下的元素返回到stk1, 这样, stk1底部多了元素x, 并且其他元素位置不变.

初始状态 清空stk1 push(5) 剩下的元素返回stk1

stk1 : 4,3,2,1 null 5 5,4,3,2,1

stk2 : null 1,2,3,4 1,2,3,4 null

但是这种push操作需要O(n)复杂度.

class MyQueue {
    Stack<Integer> stk1;
    Stack<Integer> stk2;
    /** Initialize your data structure here. */
    public MyQueue() {
        stk1 = new Stack<Integer>();
        stk2 = new Stack<Integer>();
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        while(!stk1.isEmpty())
        {
            stk2.push(stk1.pop());
        }
        stk1.push(x);
        while(!stk2.isEmpty())
        {
            stk1.push(stk2.pop());
        }
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        return stk1.pop();
    }
    
    /** Get the front element. */
    public int peek() {
        return stk1.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return stk1.isEmpty();
    }
}

/**
 * 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();
 */

2 栈模拟2

上面的方法每push一次都要有O(n)的计算量.

可以改进一下, 设置一个pushstack和popstack.

每次push都直接push到pushstack中. pop或top的时候再把pushtack中的元素全都转移过来. 由于栈的特性, 转移到popstack后自然是队列的顺序了.

这样, 所有操作的复杂度都可以降为O(1)

class MyQueue {
    Stack<Integer> pushStk;
    Stack<Integer> popStk;
    /** Initialize your data structure here. */
    public MyQueue() {
        pushStk = new Stack<Integer>();
        popStk = new Stack<Integer>();
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        pushStk.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        if(!popStk.isEmpty())
        {
            return popStk.pop();
        }
        else
        {
            while(!pushStk.isEmpty())
            {
                popStk.push(pushStk.pop());
            }
            return popStk.pop();
        }
    }
    
    /** Get the front element. */
    public int peek() {
        if(!popStk.isEmpty())
        {
            return popStk.peek();
        }
        else
        {
            while(!pushStk.isEmpty())
            {
                popStk.push(pushStk.pop());
            }
            return popStk.peek();
        }
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return pushStk.isEmpty() && popStk.isEmpty();
    }
}

/**
 * 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();
 */