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
, andis 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).
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(); */