225. Implement Stack using Queues - Easy

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.

Example:

MyStack stack = new MyStack();

stack.push(1);
stack.push(2);  
stack.top();   // returns 2
stack.pop();   // returns 2
stack.empty(); // returns false

Notes:

• You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is emptyoperations 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).

 

M1: two queues

push永远push进q1, pop -> 先pop q1 并把弹出元素放入q2直到q1只剩一个元素, 记录最后弹出的元素，不放入q2, 交换q1, q2, 返回q1最后弹出的元素

push() 　  -- time: O(1), space: O(1)

pop() 　　-- time: O(n), space: O(1)

class MyStack {
    Queue<Integer> q1;
    Queue<Integer> q2;

    /** Initialize your data structure here. */
    public MyStack() {
        q1 = new LinkedList<>();
        q2 = new LinkedList<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        q1.offer(x);
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        while(q1.size() > 1) {
            q2.offer(q1.poll());
        }
        int top = q1.poll();
        Queue<Integer> tmp = q1;
        q1 = q2;
        q2 = tmp;
        return top;
    }
    
    /** Get the top element. */
    public int top() {
        int top = 0;
        while(!q1.isEmpty()) {
            top = q1.poll();
            q2.offer(top);
        }
        Queue<Integer> tmp = q1;
        q1 = q2;
        q2 = tmp;
        return top;
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return q1.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */

 

M2: one queue

需要知道queue的大小，push先加入队尾，再rotate,  即offer(q.poll())   size - 1  次，再返回队顶值，pop/top直接操作就行

push() 　  -- time: O(n), space: O(1)

pop() 　　-- time: O(1), space: O(1)

class MyStack {
    Queue<Integer> q;

    /** Initialize your data structure here. */
    public MyStack() {
        q = new LinkedList<>();
    }
    
    /** Push element x onto stack. */
    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 and returns that element. */
    public int pop() {        
        return q.poll();
    }
    
    /** Get the top element. */
    public int top() {
        return q.peek();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return q.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */

 

转载于:https://www.cnblogs.com/fatttcat/p/10221190.html