225 Implement Stack using Queues 用队列实现栈
Description:
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 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).
题目描述:
使用队列实现栈的下列操作：
- push(x) -- 元素 x 入栈
- pop() -- 移除栈顶元素
- top() -- 获取栈顶元素
- empty() -- 返回栈是否为空
注意:
- 你只能使用队列的基本操作-- 也就是 push to back, peek/pop from front, size, 和 is empty 这些操作是合法的。
- 你所使用的语言也许不支持队列。 你可以使用 list 或者 deque（双端队列）来模拟一个队列 , 只要是标准的队列操作即可。
- 你可以假设所有操作都是有效的（例如, 对一个空的栈不会调用 pop 或者 top 操作）。
思路:
队列和栈 push操作一样, 但是队列只能从队首(front)弹出, 栈只能从栈顶(back)弹出
- push的时候将队列的元素依次弹出, push到队尾, 其他函数可以设计的一样
- 设计两个队列, 一个队列用来记录输入, 一个队列用来弹出
- push()时间复杂度O(n), 空间复杂度O(n)
- pop()时间复杂度O(1), 空间复杂度O(1)
- top()时间复杂度O(1), 空间复杂度O(1)
- empty()时间复杂度O(1), 空间复杂度O(1)
代码:
C++:
class MyStack
{
public:
/** Initialize your data structure here. */
queue<int> q;
MyStack() {}
/** Push element x onto stack. */
void push(int x)
{
q.push(x);
for (int i = 0; i < q.size() - 1; i++)
{
q.push(q.front());
q.pop();
}
}
/** Removes the element on top of the stack and returns that element. */
int pop()
{
int result = q.front();
q.pop();
return result;
}
/** Get the top element. */
int top()
{
return q.front();
}
/** Returns whether the stack is empty. */
bool empty()
{
return q.empty();
}
};
/**
* 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();
* bool param_4 = obj->empty();
*/
Java:
class MyStack {
/** Initialize your data structure here. */
public Queue<Integer> stack;
public MyStack() {
stack = new LinkedList<>();
}
/** Push element x onto stack. */
public void push(int x) {
stack.offer(x);
for (int i = 0; i < stack.size() - 1; i++) {
Integer result = stack.poll();
stack.offer(result);
}
}
/** Removes the element on top of the stack and returns that element. */
public int pop() {
return stack.poll();
}
/** Get the top element. */
public int top() {
return stack.peek();
}
/** Returns whether the stack is empty. */
public boolean empty() {
return stack.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();
*/
Python:
class MyStack:
def __init__(self):
"""
Initialize your data structure here.
"""
self.input = []
self.output = []
def push(self, x: int) -> None:
"""
Push element x onto stack.
"""
self.input.append(x)
while self.output:
self.input.append(self.output.pop(0))
self.output, self.input = self.input, []
def pop(self) -> int:
"""
Removes the element on top of the stack and returns that element.
"""
return self.output.pop(0)
def top(self) -> int:
"""
Get the top element.
"""
return self.output[0]
def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
return not self.output
# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()
