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.

来源： http://www.lintcode.com/en/problem/implement-queue-by-two-stacks/

分析

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

public  class  Queue {      private  Stack<Integer> stack1 =  new  Stack<Integer>();      private  Stack<Integer> stack2 =  new  Stack<Integer>();        public  Queue() {         // do initialization if necessary      }            public  void  push( int  element) {          // write your code here          while (!stack1.empty()){              stack2.push(stack1.pop());          }          stack1.push(element);          while (!stack2.empty()){              stack1.push(stack2.pop());          }      }        public  int  pop() {          // write your code here          if (!stack1.empty())              return  stack1.pop();          else              return  - 1 ;      }        public  int  top() {          // write your code here          if (!stack1.empty())              return  stack1.peek();          else              return  - 1 ;      } }

null

转载于:https://www.cnblogs.com/zhxshseu/p/9415037aea8647d4817529ba5f792bde.html