Cracking coding interview(3.5)使用2个堆栈实现一个队列

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3.5 Implement a MyQueue class which implements a queue using two stacks.

explanation:

1.MyQueue实现原理：当MyQueue需要offer(add)一个元素时，直接添加到堆栈s1中即可，若要poll(remove)一个元素时，则需要借助堆栈s2，将s1中元素pop并且push到s2中，此时s2栈顶元素即是对头元素。之后再将s2中元素pop并且push回s1中。

2.MyQueue2针对MyQueue的优化。堆栈s2中的元素不再回到s1.堆栈s1用于存放最新的push的元素，堆栈s2用于将堆栈s1中的现有元素反序，这样最先入队的几个元素都会在s2中。当MyQueue队列要弹出一个元素时，只需要从s2中弹出元素即可，若s2为空，则将s1中的最新的元素再次pop->push到s2。如此往复。

import java.util.Stack;class MyQueue{private Stack<Integer> s1 = new Stack<Integer>();private Stack<Integer> s2 = new Stack<Integer>();//time complexity:O(1), space complexity:O(1)public boolean offer(int val){s1.push(val);return true;}//time complexity:O(n) space complexity:O(n)public int poll(){if(empty())return Integer.MIN_VALUE;else{while(!s1.empty())s2.push(s1.pop());int val = s2.pop().intValue();while(!s2.empty())s1.push(s2.pop());return val;}}public boolean empty(){return s1.empty();}}//improvement for MyQueueclass MyQueue2{Stack<Integer> s1 = new Stack<Integer>();Stack<Integer> s2 = new Stack<Integer>();public boolean offer(int val){s1.push(val);return true;}//time complexity gradually reduced, most cases is O(1)public int poll(){if(s2.empty())while(!s1.empty())s2.push(s1.pop());if(s2.empty())return Integer.MIN_VALUE;elsereturn s2.pop().intValue();}public boolean empty(){if(s1.empty() && s2.empty())return true;elsereturn false;}}public class Solution{public static void main(String[] args){//test for MyQueueMyQueue mq = new MyQueue();int i=0;for(;i < 20;i++)mq.offer(i);while(--i > 10 && !mq.empty())System.out.print(mq.poll()+" ");System.out.println();for(i=40;i < 50;i++)mq.offer(i);while(!mq.empty())System.out.print(mq.poll()+" ");System.out.println();//test for MyQueue2MyQueue2 mq2 = new MyQueue2();for(i=100;i < 102;i++)mq2.offer(i);while(i-- != 101 &&!mq2.empty())System.out.print(mq2.poll()+" ");for(i=102;i < 120;i++)mq2.offer(i);while(!mq2.empty())System.out.print(mq2.poll()+" ");System.out.println();}}

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