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Implement Stack using Queues

描述

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.

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).

分析

可以用两个队列,qtmpq存放元素,tmp用来作中转。

  • push(x),先将x push 到tmp,然后把q中的元素全部弹出来,存入tmp,最后切换qtmp
  • pop(),直接将q的队首元素弹出来即可

该算法push的算法复杂度是O(n), pop的算法复杂度O(1)

另个一个方法是,让popO(n), pushO(1),思路很类似,就不赘述了。

代码

// Implement Stack using Queues
class MyStack {
// Push element x onto stack.
// Time Complexity O(n)
public void push(int x) {
tmp.offer(x);
while (!q.isEmpty()) {
final int e = q.poll();
tmp.offer(e);
}
// swap q and tmp
Queue<Integer> temp = tmp;
tmp = q;
q = temp;
}

// Removes the element on top of the stack.
// Time Complexity O(1)
public void pop() {
q.poll();
}

// Get the top element.
public int top() {
return q.peek();
}

// Return whether the stack is empty.
public boolean empty() {
return q.isEmpty();
}

private Queue<Integer> q = new LinkedList<>();
private Queue<Integer> tmp = new LinkedList<>();
}

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