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
1 public class Queue { 2 private Stack<Integer> stack1; 3 private Stack<Integer> stack2; 4 5 public Queue() { 6 // do initialization if necessary 7 stack1 = new Stack<Integer>(); 8 stack2 = new Stack<Integer>(); 9 } 10 11 public void push(int element) { 12 stack1.push(element); 13 } 14 15 public int pop() { 16 if (stack2.size() != 0) { 17 return stack2.pop(); 18 } 19 20 while (stack1.size() != 0) { 21 stack2.push(stack1.pop()); 22 } 23 24 if (stack2.size() != 0) { 25 return stack2.pop(); 26 } 27 return -1; 28 } 29 30 public int top() { 31 32 if (stack2.size() != 0) { 33 return stack2.peek(); 34 } 35 while (stack1.size() != 0) { 36 stack2.push(stack1.pop()); 37 } 38 39 if (stack2.size() != 0) { 40 return stack2.peek(); 41 } 42 return -1; 43 } 44 }
原文：http://www.cnblogs.com/beiyeqingteng/p/5683314.html