LeetCode--Implement Stack using Queues

Problem:

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.

Analysis:

【解法一：用两个队列，push: O(1)，pop: O(n)，top: O(n)】

用两个队列q1，q2实现一个栈。push时把新元素添加到q1的队尾。pop时把q1中除最后一个元素外逐个添加到q2中，然后pop掉q1中的最后一个元素，然后注意记得q1和q2，以保证我们添加元素时始终向q1中添加。top的道理类似。

【解法二：用两个队列，push: O(n)，pop: O(1)，top: O(1)】

所有元素都倒序保存在q1中，即后添加的元素在q1的最前端，如何做到呢？每次push时，把新元素放到空的q2，然后把q1中元素逐个添加到q2的队尾，最后交换q1和q2。这样q1队首的元素就是最后添加的元素，pop和top直接返回q1队首的元素就好。

Anwser1:

class MyStack {
Queue<Integer> in =  new LinkedList<Integer>();
Queue<Integer> out = new LinkedList<Integer>();
// Push element x onto stack.
public void push(int x) {
in.add(x);
}

// Removes the element on top of the stack.
public void pop() {
while(in.size()>1) out.add(in.poll());
in.poll();
Queue<Integer> q=in;
in = out;
out = q;
}

// Get the top element.
public int top() {
while(in.size()>1) out.add(in.poll());
int x = in.poll();
out.add(x);
Queue<Integer> q=in;
in = out;
out = q;
return x;
}

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

Anwser2:

class MyStack {
private Queue<Integer> q1 = new LinkedList<>();
private Queue<Integer> q2 = new LinkedList<>();

// Push element x onto stack.
public void push(int x) {
q2.offer(x);
while (!q1.isEmpty()) {
q2.offer(q1.poll());
}

Queue tmp = q1;
q1 = q2;
q2 = tmp;
}

// Removes the element on top of the stack.
public void pop() {
q1.poll();
}

// Get the top element.
public int top() {
return q1.peek();
}

// Return whether the stack is empty.
public boolean empty() {
return q1.isEmpty();
}
}