225. Implement Stack using Queues

题目:

Implement the following operations of a stack using queues.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• empty() -- Return whether the stack is empty.

Notes:

• You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid.
• Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
• You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

 

Update (2015-06-11):
The class name of the Java function had been updated to MyStack instead of Stack.

链接： http://leetcode.com/problems/implement-stack-using-queues/

题解：

一开始想使用两个Queue，来回倒腾一下就可以得到结果，但这样基本每个op都是O(n)。后来看了Discuss，发现一个Queue就可以，然后只有Push()的Complexity - O(n)，应该算是optimal了。不过仔细想一想其实Push最好能是O(1)，因为这个用得应该最频繁。 下面做法主要就是在push的时候把当前值放在队尾，以前的值我们重新放在队尾。

Time Complexity - Push - O(n)， Pop - O(1)， Peek - O(1)，  isEmpty- O(1)

class MyStack {
    // Push element x onto stack.
    Queue<Integer> q;
    
    public MyStack() {
        q = new LinkedList<>();
    }
    
    public void push(int x) {
        q.offer(x);
        
        for(int i = 0; i < q.size() - 1; i++) {
            q.offer(q.peek());
            q.poll();
        }
    }

    // Removes the element on top of the stack.
    public void pop() {
        q.poll();
    }

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

    // Return whether the stack is empty.
    public boolean empty() {
        return q.size() == 0;
    }
}

 

二刷:

使用一个queue来求解，每次push的时候把q头部poll出的数字再加回到尾部

Java:

Time Complexity - Push - O(n)， Pop - O(1)， Peek - O(1)，  isEmpty- O(1)

class MyStack {
    // Push element x onto stack.
    Queue<Integer> q = new LinkedList();
    
    public void push(int x) {
        q.offer(x);
        for (int i = 0; i < q.size() - 1; i++) {
            q.offer(q.poll());
        }
    }

    // Removes the element on top of the stack.
    public void pop() {
        q.poll();
    }

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

    // Return whether the stack is empty.
    public boolean empty() {
        return q.size() == 0;
    }
}

 

三刷:

Java:

class MyStack {
    private Queue<Integer> q = new LinkedList<>();
    
    // Push element x onto stack.
    public void push(int x) {
        q.offer(x);
        for (int i = q.size() - 2; i >= 0; i--) {
            q.offer(q.poll());
        }
    }

    // Removes the element on top of the stack.
    public void pop() {
        q.poll();
    }

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

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

 

Update: 不过实际运行速度比较慢啊

Time Complexity - Push - O(n)， Pop - O(1)， Peek - O(1)，  isEmpty- O(1)

class MyStack {
    Queue<Integer> q = new LinkedList<>();
    // Push element x onto stack.
    public void push(int x) {
        q.offer(x);
        int len = q.size();
        while (len > 1) {
            q.offer(q.poll());
            len--;
        }
    }

    // Removes the element on top of the stack.
    public void pop() {
        q.poll();
    }

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

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

 

 

Reference:

https://leetcode.com/discuss/46975/a-simple-c-solution

https://leetcode.com/discuss/40202/only-push-others-using-queue-combination-shared-solutions 

https://leetcode.com/discuss/84233/solutions-about-which-utilizes-queue-others-utilize-queues

https://leetcode.com/discuss/39839/o-1-purely-with-queues