【LeetCode】Implement Stack using Queues 解题报告

【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).

【解法一：用两个队列，push: O(1)，pop: O(n)，top: O(n)】
用两个队列q1，q2实现一个栈。push时把新元素添加到q1的队尾。pop时把q1中除最后一个元素外逐个添加到q2中，然后pop掉q1中的最后一个元素，然后注意记得q1和q2，以保证我们添加元素时始终向q1中添加。top的道理类似。

class MyStack {
    private Queue<Integer> q1 = new LinkedList<>();
    private Queue<Integer> q2 = new LinkedList<>();
    
    // Push element x onto stack.
    public void push(int x) {
        q1.offer(x);
    }

    // Removes the element on top of the stack.
    public void pop() {
        while (q1.size() > 1) {
            q2.offer(q1.poll());
        }
        q1.poll();
        
        Queue tmp = q1;
        q1 = q2;
        q2 = tmp;
    }

    // Get the top element.
    public int top() {
        while (q1.size() > 1) {
            q2.offer(q1.poll());
        }
        int top = q1.peek();
        q2.offer(q1.poll());
        
        Queue tmp = q1;
        q1 = q2;
        q2 = tmp;
        
        return top;
    }

    // Return whether the stack is empty.
    public boolean empty() {
        return q1.isEmpty();
    }
}

【解法二：用两个队列，push: O(n)，pop: O(1)，top: O(1)】

所有元素都倒序保存在q1中，即后添加的元素在q1的最前端，如何做到呢？每次push时，把新元素放到空的q2，然后把q1中元素逐个添加到q2的队尾，最后交换q1和q2。这样q1队首的元素就是最后添加的元素，pop和top直接返回q1队首的元素就好。

class MyStack {
    private Queue<Integer> q1 = new LinkedList<>();
    private Queue<Integer> q2 = new LinkedList<>();
    
    // Push element x onto stack.
    public void push(int x) {
        q2.offer(x);
        while (!q1.isEmpty()) {
            q2.offer(q1.poll());
        }
        
        Queue tmp = q1;
        q1 = q2;
        q2 = tmp;
    }

    // Removes the element on top of the stack.
    public void pop() {
        q1.poll();
    }

    // Get the top element.
    public int top() {
        return q1.peek();
    }

    // Return whether the stack is empty.
    public boolean empty() {
        return q1.isEmpty();
    }
}

【解法三：一个队列，push: O(1)，pop: O(n)，top: O(n)】

push时直接添加到队尾就好。pop和top时，把队列除最后一个元素外，逐个循环添加到队列的尾部。

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

    // Removes the element on top of the stack.
    public void pop() {
        int size = q.size();
        for (int i = 1; i < size; i++) {
            q.offer(q.poll());
        }
        q.poll();
    }

    // Get the top element.
    public int top() {
        int size = q.size();
        for (int i = 1; i < size; i++) {
            q.offer(q.poll());
        }
        int top = q.peek();
        q.offer(q.poll());
        return top;
    }

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

其实三种实现，大同小异，无非是队列中元素的不停pop与push以得到最后一个元素。