leetcode解題之225 # Implement Stack using Queuest Java版 (用兩個佇列實現一個棧)
阿新 • • 發佈:2019-01-11
225. Implement Stack using Queues
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.
- 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).
可以用兩個Queue來輪流儲存資料,當執行pop,top等指令時,因為要返回Queue中最後一個element,所以把之前的所有elements壓到另外一個空queue中,剩下唯一一個element的時候,實現poll返回。依次交替
注意事項:pop,top等取出行為,要先檢查兩個queue是不是都為空
import java.util.LinkedList; import java.util.Queue; public class MyStack { Queue<Integer> q1 = new LinkedList<Integer>(); Queue<Integer> q2 = new LinkedList<Integer>(); /** Initialize your data structure here. */ public MyStack() { } /** Push element x onto stack. */ public void push(int x) { if (q2.isEmpty()) { q1.offer(x); } else q2.offer(x); } /** Removes the element on top of the stack and returns that element. */ public int pop() { while (!q2.isEmpty()) { if (q2.size() == 1) return q2.poll(); q1.offer(q2.poll()); } while (!q1.isEmpty()) { if (q1.size() == 1) return q1.poll(); q2.offer(q1.poll()); } return -1; } /** Get the top element. */ public int top() { while (!q2.isEmpty()) { if (q2.size() == 1) { int x = q2.peek(); q1.offer(q2.poll()); return x; } q1.offer(q2.poll()); } while (!q1.isEmpty()) { if (q1.size() == 1) { int x = q1.peek(); q2.offer(q1.poll()); return x; } q2.offer(q1.poll()); } return -1; } /** Returns whether the stack is empty. */ public boolean empty() { return q1.isEmpty() && q2.isEmpty(); } }