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(LeetCode)用兩個棧實現一個佇列

LeetCode上面的一道題目,原文如下:

Implement the following operations of a queue using stacks.

  • push(x) -- Push element x to the back of queue.
  • pop() -- Removes the element from in front of queue.
  • peek() -- Get the front element.
  • empty() -- Return whether the queue is empty.
Notes:
  • You must use only standard operations of a stack -- which means only push to top
    peek/pop from topsize, and is empty operations are valid.
  • Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
  • You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
我的思路是建立兩個棧S1和S2,入隊時,將元素壓入S1,出隊時,如果棧S2中的元素個數為0,則將S1中的元素一個個壓入S2,並彈出最後壓入的那個元素,如果棧S2中的元素個數不為0,則直接彈出S2中的頂元素。

程式碼如下:

class MyQueue {
    // Push element x to the back of queue.
    Stack<Integer> stack1 = new Stack<Integer>();
	Stack<Integer> stack2 = new Stack<Integer>();
    public void push(int x) {
        stack1.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        if(stack2.size()==0)
		{
		        int m = stack1.size();
				for(int i=0;i<m;i++)
				{
					stack2.push(stack1.pop());
				}
		}
			stack2.pop();
    }

    // Get the front element.
    public int peek() {
        if(stack2.size()==0)
		{
		    int m = stack1.size();
			for(int i=0;i<m;i++)
			{
				stack2.push(stack1.pop());
			}
		}
			return stack2.peek();
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return stack1.size()==0&&stack2.size()==0;
    }
}