Implement Queue by Two Stacks
As the title described, you should only use two stacks to implement a queue‘s actions.
The queue should support push(element), pop() and top() where pop is pop the first(a.k.a front) element in the queue.
Both pop and top methods should return the value of first element.
push(1)
pop() // return 1
push(2)
push(3)
top() // return 2
pop() // return 2
implement it by two stacks, do not use any other data structure and push, pop and top should be O(1) by AVERAGE.
public class Queue { private Stack<Integer> stack1; private Stack<Integer> stack2; public Queue() { // do initialization if necessary stack1=new Stack<Integer>(); stack2=new Stack<Integer>(); } public void push(int element) { // write your code here stack1.push(element); } public int pop() { // write your code here if(stack2.isEmpty()) { while(!stack1.isEmpty()) { stack2.push(stack1.pop()); } } return stack2.pop(); } public int top() { // write your code here if(stack2.isEmpty()) { while(!stack1.isEmpty()) { stack2.push(stack1.pop()); } } return stack2.peek(); } }
[lintcode medium]Implement Queue by Two Stacks
原文：http://www.cnblogs.com/kittyamin/p/5104264.html