Implement the following operations of a queue using stacks.
- push(x) -- Push element x to the back of queue.
- pop() -- Removes the element from in front of queue.
- peek() -- Get the front element.
- empty() -- Return whether the queue is empty.
Notes:
- You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
- Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
一刷
题解：这题跟225的区别是，
用两个stack, input, output
如果仅仅add, 则push到input中去。
如果pop/peek, 如果output为空，则将input中的num全部pop, push到output, 否则只从output中取，不动input
原因是，如果push(1), push(2), peek(), push(3), peek(), 这样就可以保证顺序了。
public class MyStack {
private Queue<Integer> q1;
private Queue<Integer> q2;
/** Initialize your data structure here. */
public MyStack() {
q1 = new LinkedList<>();
q2 = new LinkedList<>();
}
/** Push element x onto stack. */
public void push(int x) {
if(q1.isEmpty()){
q1.add(x);
for(int i=0; i<q2.size(); i++){
q1.add(q2.poll());
}
}
else{
q2.add(x);
for(int i=0; i<q1.size(); i++){
q2.add(q1.poll());
}
}
}
/** Removes the element on top of the stack and returns that element. */
public int pop() {
if(q1.isEmpty()) return q2.poll();
else return q1.poll();
}
/** Get the top element. */
public int top() {
if(q1.isEmpty()) return q2.peek();
else return q1.peek();
}
/** Returns whether the stack is empty. */
public boolean empty() {
return q1.isEmpty() && q2.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();
*/
