LeetCode(23)-栈实现队列

LeetCode(23)-Implement Queue using Stacks(栈实现队列)

题目:

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 top, size, 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).

思路:

  • 这道题的意思是用堆栈(stack)来实现队列(quene)
  • 可以用两个stack,s1和s2来实现,s1用来整体的存储,s2用来缓冲倒叙,首先判断s2是否为空,为空,把s1弹出到s2,否则s2.pop()

代码:

class MyQueue {
    Stack<Integer> s1 = new Stack<>();
    Stack<Integer> s2 = new Stack<>();
    // Push element x to the back of queue.
    public void push(int x) {
        s1.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        if(!s2.isEmpty()){
            s2.pop();
        }else{
            if(s1.isEmpty()){
                return;
            }else{
                while(!s1.isEmpty()){
                    int tmp = s1.pop();
                    s2.push(tmp);
                }
                s2.pop();
            }
        }
        
    }

    // Get the front element.
    public int peek() {
        if(!s2.isEmpty()){
            return s2.peek();
        }else{
            if(s1.isEmpty()){
                return -1;
            }else{
                while(!s1.isEmpty()){
                    int tmp = s1.pop();
                    s2.push(tmp);
                }
                return s2.peek();
            }
        }
        
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return(s1.isEmpty() && s2.isEmpty());
        
    }
}

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    原文作者:牧童遥指2000
    原文地址: https://www.jianshu.com/p/4edfbdcb9df0
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