232. Implement Queue using Stacks(java)

Implement the following operations of a queue using stacks.

• push(x) -- Push element x to the back of queue.
• pop() -- Removes the element from in front of queue.
• peek() -- Get the front element.
• empty() -- Return whether the queue is empty.

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, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
• You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

用栈实现队列

思路一：在进队列 的时候，把栈s1逆序压入另一个辅助栈s2，再逆序入s1；出队列的时候直接出栈s1。

public class MyQueue {
    
    private int front;
    Stack s1=new Stack<>();
    Stack s2=new Stack<>();
    
    /** Initialize your data structure here. */
    public MyQueue() {
        

    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        if(s1.isEmpty())
            front=x;
        while(!s1.isEmpty()){
            s2.push(s1.pop());
        }
        s2.push(x);
        while(!s2.isEmpty()){
            s1.push(s2.pop());
        }
        
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        int tmp=s1.pop();
        if(!s1.isEmpty()) 
            front=s1.peek();
        return tmp;
        
    }
    
    /** Get the front element. */
    public int peek() {
        return front;
        
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return s1.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();
 */

思路二：进队列的时候正常入栈s1；出队列的时候把栈s1逆序放在另外一个栈s2，出栈s2。

官网solution：

Approach #1 (Two Stacks) Push - $O(n)$ per operation, Pop - $O(1)$ per operation.

Algorithm

Push

A queue is FIFO (first-in-first-out) but a stack is LIFO (last-in-first-out). This means the newest element must be pushed to the bottom of the stack. To do so we first transfer all s1 elements to auxiliary stack s2. Then the newly arrived element is pushed on top of s2 and all its elements are popped and pushed to s1.

Figure 1. Push an element in queue

Java

private int front;

public void push(int x) {
    if (s1.empty())
        front = x;
    while (!s1.isEmpty())
        s2.push(s1.pop());
    s2.push(x);
    while (!s2.isEmpty())
        s1.push(s2.pop());
}

Complexity Analysis

• Time complexity : $O(n)$.

Each element, with the exception of the newly arrived, is pushed and popped twice. The last inserted element is popped and pushed once. Therefore this gives $4n+2$ operations where $n$ is the queue size. The push and pop operations have $O(1)$ time complexity.

• Space complexity : $O(n)$. We need additional memory to store the queue elements

Pop

The algorithm pops an element from the stack s1, because s1 stores always on its top the first inserted element in the queue. The front element of the queue is kept as front.

Figure 2. Pop an element from queue

Java

// Removes the element from the front of queue.
public void pop() {
    s1.pop();
    if (!s1.empty())
        front = s1.peek();
}

Complexity Analysis

• Time complexity : $O(1)$.
• Space complexity : $O(1)$.

Empty

Stack s1 contains all stack elements, so the algorithm checks s1 size to return if the queue is empty.

// Return whether the queue is empty.
public boolean empty() {
    return s1.isEmpty();
}

Time complexity : $O(1)$.

Space complexity : $O(1)$.

Peek

The front element is kept in constant memory and is modified when we push or pop an element.

// Get the front element.
public int peek() {
  return front;
}

Time complexity : $O(1)$. The front element has been calculated in advance and only returned in peek operation.

Space complexity : $O(1)$.

---

Approach #2 (Two Stacks) Push - $O(1)$ per operation, Pop - Amortized $O(1)$ per operation.

Algorithm

Push

The newly arrived element is always added on top of stack s1 and the first element is kept as front queue element

Figure 3. Push an element in queue

Java

private Stack<Integer> s1 = new Stack<>();
private Stack<Integer> s2 = new Stack<>();

// Push element x to the back of queue.
public void push(int x) {
    if (s1.empty())
        front = x;
    s1.push(x);
}

Complexity Analysis

• Time complexity : $O(1)$.

Аppending an element to a stack is an O(1) operation.

• Space complexity : $O(n)$. We need additional memory to store the queue elements

Pop

We have to remove element in front of the queue. This is the first inserted element in the stack s1 and it is positioned at the bottom of the stack because of stack'sLIFO (last in - first out) policy. To remove the bottom element from s1, we have to pop all elements from s1 and to push them on to an additional stack s2, which helps us to store the elements of s1 in reversed order. This way the bottom element of s1 will be positioned on top of s2 and we can simply pop it from stacks2. Once s2 is empty, the algorithm transfer data from s1 to s2 again.

Figure 4. Pop an element from stack

Java

// Removes the element from in front of queue.
public void pop() {
    if (s2.isEmpty()) {
        while (!s1.isEmpty())
            s2.push(s1.pop());
    }
    s2.pop();    
}

Complexity Analysis

• Time complexity: Amortized $O(1)$, Worst-case $O(n)$.

In the worst case scenario when stack s2 is empty, the algorithm pops $n$ elements from stack s1 and pushes $n$ elements to s2, where $n$ is the queue size. This gives$2n$ operations, which is $O(n)$. But when stack s2 is not empty the algorithm has $O(1)$ time complexity. So what does it mean by Amortized $O(1)$? Please see the next section on Amortized Analysis for more information.

• Space complexity : $O(1)$.

Amortized Analysis

Amortized analysis gives the average performance (over time) of each operation in the worst case. The basic idea is that a worst case operation can alter the state in such a way that the worst case cannot occur again for a long time, thus amortizing its cost.

Consider this example where we start with an empty queue with the following sequence of operations applied:

push_1, push_2, \ldots, push_n, pop_1,pop_2 \ldots, pop_npush​1​​,push​2​​,…,push​n​​,pop​1​​,pop​2​​…,pop​n​​

The worst case time complexity of a single pop operation is $O(n)$. Since we have $n$ pop operations, using the worst-case per operation analysis gives us a total of O(n^2)O(n​2​​) time.

However, in a sequence of operations the worst case does not occur often in each operation - some operations may be cheap, some may be expensive. Therefore, a traditional worst-case per operation analysis can give overly pessimistic bound. For example, in a dynamic array only some inserts take a linear time, though others - a constant time.

In the example above, the number of times pop operation can be called is limited by the number of push operations before it. Although a single pop operation could be expensive, it is expensive only once per n times (queue size), when s2 is empty and there is a need for data transfer between s1 and s2. Hence the total time complexity of the sequence is : n (for push operations) + 2*n (for first pop operation) + n - 1 ( for pop operations) which is $O(2\ast n)$.This gives $O(2n/2n)$ = $O(1)$ average time per operation.

Empty

Both stacks s1 and s2 contain all stack elements, so the algorithm checks s1 and s2 size to return if the queue is empty.

// Return whether the queue is empty.
public boolean empty() {
    return s1.isEmpty() && s2.isEmpty();
}

Time complexity : $O(1)$.

Space complexity : $O(1)$.

Peek

The front element is kept in constant memory and is modified when we push an element. When s2 is not empty, front element is positioned on the top of s2

// Get the front element.
public int peek() {
    if (!s2.isEmpty()) {
            return s2.peek();
    }
    return front;
}

Time complexity : $O(1)$.

The front element was either previously calculated or returned as a top element of stack s2. Therefore complexity is $O(1)$

Space complexity : $O(1)$.