题目描述
Implement the following operations of a queue using stacks.
- push(x) – Push element x to the back of queue.
- pop() – Removes the element from in front of queue.
- peek() – Get the front element.
- empty() – Return whether the queue is empty.
Notes:
- You must use only standard operations of a stack – which means only push to top, peek/pop from top, size, and is empty operations are valid.
- Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
分析
参考:LeetCode 225 Implement Stack using Queues
非常经典的题目,定义两个栈模拟队列。
- push向stack1
- pop从stack2,如果stack2为空,先将stack1的元素放入stack2
代码
class MyQueue {
Stack<Integer> stack1 = new Stack<Integer>();
Stack<Integer> stack2 = new Stack<Integer>();
// Push element x to the back of queue.
public void push(int x) {
stack1.push(x);
}
// Removes the element from in front of queue.
public void pop() {
if (stack2.isEmpty()) {
if (stack1.isEmpty()) {
throw new IllegalStateException();
}
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
}
stack2.pop();
}
// Get the front element.
public int peek() {
if (stack2.isEmpty()) {
if (stack1.isEmpty()) {
throw new IllegalStateException();
}
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
}
return stack2.peek();
}
// Return whether the queue is empty.
public boolean empty() {
if (stack1.isEmpty() && stack2.isEmpty()) {
return true;
}
return false;
}
}