Implement Queue using Two Stacks
阿新 • • 發佈：2018-02-21
som when tcl caller shuffle make com isempty use
/* how to use two stacks to implement the queue: offer, poll, peek,size, isEmpty offer(3) offer(2) poll() offer(1) peek() offer(6) poll() poll() 3 2 2 1 in (3 2) (1 6) out (2) (3) 6 (1) stack1(in): is the only stack to store new elements when adding a new element into the queue stack2(out): is the only stack to pop old element out of the queue. when stack2 is empty, we move all data from stack1(in) to stack2(out) if any **/ public class QueueUseStack { private Deque<Integer> in ; private Deque<Integer> out; public QueueUseStack(){ in = new LinkedList<>(); out = new LinkedList<>(); } //offer: always goes to the in: o(1) public void offer(int val){ in.push(val); }/*poll: * if the out has items, then pop; * else shuffle the in to out, then pop * corner cases: make the return type Integer, so the caller could do some checks * before pop, if the out still empty, return null * */ public Integer poll(){ if (out.isEmpty()){//shuffle shuffle(); } if (out.isEmpty()) return null ; return out.pop(); } //the only chance we need to shuffle is when out is empty and the caller is invoking poll or peek private void shuffle(){ while (!in.isEmpty()){ out.push(in.pop()); } } //peek public Integer peek(){ if (out.isEmpty()){ //shuffle shuffle(); } if (out.isEmpty()) return null ; return out.peek(); } //size: o(1) public int size(){ return in.size() + out.size(); } //isEmpty: o(1) public boolean isEmpty(){ return in.isEmpty() && out.isEmpty(); } } class TestClass{ public static void main(String[] args){ QueueUseStack queueUseStack = new QueueUseStack(); //offer(3) offer(2) poll() offer(1) peek() offer(6) poll() poll() queueUseStack.offer(3); queueUseStack.offer(2); System.out.println(queueUseStack.poll()); //3 queueUseStack.offer(1); System.out.println(queueUseStack.peek()); //2 queueUseStack.offer(6); System.out.println(queueUseStack.poll()); //2 System.out.println(queueUseStack.poll()); //1 } }
time complexity:
offer(), size(), isEmpty(): o(1)
for poll and peek:
worst case: offer n times, peek/poll only once: o(n)
average case: offer n times, peek/poll n times
2n: offer n times + peek/poll n times
n: shuffle n times: totally there are n items to shuffle
/n: average time complexity for each peek/poll
amortized time complexity: o((2n+n)/n) = o(1)
Implement Queue using Two Stacks