(LeetCode)两个队列来实现一个栈
原题例如以下:
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.
- You must use only standard operations of a queue -- which means only
push to back
,peek/pop from front
,size
, andis 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).
Java实现代码例如以下:
class MyStack { LinkedList<Integer> queue1 = new LinkedList<Integer>(); LinkedList<Integer> queue2 = new LinkedList<Integer>(); // Push element x onto stack. public void push(int x) { if(queue1.size()==0&&queue2.size()==0) { queue1.offer(x); } else if(queue1.size()==0) { queue2.offer(x); } else { queue1.offer(x); } } // Removes the element on top of the stack. public void pop() { if(queue1.size()!= 0) { int length = queue1.size(); for(int i =0;i<length-1;i++) { queue2.offer(queue1.poll()); } queue1.poll(); } else { int length = queue2.size(); for(int i =0;i<length-1;i++) { queue1.offer(queue2.poll()); } queue2.poll(); } } // Get the top element. public int top() { if(queue1.size()!= 0) { int length = queue1.size(); for(int i =0;i<length-1;i++) { queue2.offer(queue1.poll()); } int result = queue1.element(); queue2.offer(queue1.poll()); return result; } else { int length = queue2.size(); for(int i =0;i<length-1;i++) { queue1.offer(queue2.poll()); } int result = queue2.element(); queue1.offer(queue2.poll()); return result; } } // Return whether the stack is empty. public boolean empty() { return queue1.size()==0&&queue2.size()==0; } }