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

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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;     } }

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