【leetcode】225. Implement Stack using Queues
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题目要求:
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).
- 即让你用队列实现栈
- 基本思想如下:
- 入栈的操作:如果第一个队列不为空,就把元素加入第一个队列,否则,就加入第二个队列
- 弹栈的操作:如果第一个队列不为空,就把第一个队列除最后一个元素之外的元素全部出队,然后加入第二个队列,这样第一个队列就只剩下最后进队的元素,也就是栈顶元素。比如:把1,2,3,4,5依次入栈,用队列实现时,我们会把1,2,3,4,5依次加入第一个队列,当要求弹栈时,需要弹出的应该是5这个元素,所以我们把第一个队列中的1,2,3,4依次出队,然后加入第二个队列,这样第一个队列就只剩下元素5,弹出即可。
- top和pop的思想类似
class MyStack { public Queue<Integer> queue1 = new LinkedList<Integer>(); public Queue<Integer> queue2 = new LinkedList<Integer>(); private int size; // Push element x onto stack. public void push(int x) { if(empty()||!queue1.isEmpty()) { queue1.offer(x); }else { queue2.offer(x); } size++; } // Removes the element on top of the stack. public void pop() { if(!queue1.isEmpty()) { while(queue1.size()>1) { queue2.offer(queue1.poll()); } queue1.poll(); }else { while(queue2.size()>1) { queue1.offer(queue2.poll()); } queue2.poll(); } size--; } // Get the top element. public int top() { if(!queue1.isEmpty()) { while(queue1.size()>1) { queue2.offer(queue1.poll()); } int k=queue1.poll(); queue2.offer(k); return k; }else { while(queue2.size()>1) { queue1.offer(queue2.poll()); } int k=queue2.poll(); queue1.offer(k); return k; } } // Return whether the stack is empty. public boolean empty() { return size==0?true:false; }}
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