Lintcode: Implement Queue by Stacks 解题报告

Implement Queue by Stacks

原题链接 ： http://lintcode.com/zh-cn/problem/implement-queue-by-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.

样例

For push(1), pop(), push(2), push(3), top(), pop(), you should return 1, 2 and 2

挑战

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 Solution {
private Stack<Integer> stack1;
private Stack<Integer> stack2;

public Solution() {
// 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
// write your code here
if (stack2.isEmpty()) {
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
}

return stack2.peek();
}
}

View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/lintcode/stack/StackQueue.java