232. Implement Queue using Stacks

1. Question

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

• You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

2. Examples

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

3. Constraints

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, peek, and empty.
• All the calls to pop and peek are valid.

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

4. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/implement-queue-using-stacks 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

5. Solutions

class MyQueue {

  Stack<Integer> sa;
  Stack<Integer> sb;

  public MyQueue() {
    sa = new Stack<>();
    sb = new Stack<>();
  }

  public void push(int x) {
    while (!sb.isEmpty()) {
      sa.push(sb.pop());
    }
    sa.push(x);
  }

  public int pop() {
    return getFirst(true);
  }

  public int peek() {
    return getFirst(false);
  }

  public boolean empty() {
    while (!sb.isEmpty()) {
      sa.push(sb.pop());
    }
    return sa.isEmpty();
  }

  private int getFirst(boolean needPop) {
    while (!sa.isEmpty()) {
      sb.push(sa.pop());
    }
    int a = needPop ? sb.pop() : sb.peek();
    return a;
  }
}