Problem: 232. 用栈实现队列

思路

👨‍🏫 路飞题解

复杂度

时间复杂度:

添加时间复杂度, 示例： $O(n)$

空间复杂度:

添加空间复杂度, 示例： $O(n)$

💖 朴素版

class MyQueue {Stack<Integer> s1;Stack<Integer> s2;public MyQueue() {s1 = new Stack<>();s2 = new Stack<>();}public void push(int x) {while(!s1.isEmpty()){s2.push(s1.pop());}s2.push(x);while(!s2.isEmpty()){s1.push(s2.pop());}}public int pop() {return s1.pop();}public int peek() {return s1.peek();}public boolean empty() {return s1.isEmpty();}
}/*** Your MyQueue object will be instantiated and called as such:* MyQueue obj = new MyQueue();* obj.push(x);* int param_2 = obj.pop();* int param_3 = obj.peek();* boolean param_4 = obj.empty();*/

💖 优化版

class MyQueue {private Stack<Integer> A;private Stack<Integer> B;public MyQueue() {A = new Stack<>();B = new Stack<>();}public void push(int x) {A.push(x);}public int pop() {int peek = peek();B.pop();return peek;}public int peek() {if (!B.isEmpty()) return B.peek();if (A.isEmpty()) return -1;while (!A.isEmpty()){B.push(A.pop());}return B.peek();}public boolean empty() {return A.isEmpty() && B.isEmpty();}
}