leetcode_Implement Queue using Stacks

标签：implement queue usin   stack   leetcode   

描述：

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.

• You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is emptyoperations 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：

1.用栈来实现一个队列，也就是用后进先出的栈实现先进先出的队列

2.这个还是很难想的，但总之还是比用队列来实现栈容易想，大概就是用两个栈stack1和stack2来模拟队列

3.所有的元素都从stack1进栈，所有元素都从stack2出栈，当stack2为空的时候，将stack1中的所有元素出栈并全部push到stack2中去，然后从stack2出栈

4.由于栈是后进先出的，两次后进先出的操作就实现了队列的功能

代码：

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

    // Removes the element from in front of queue.
    public void pop() {
        if(!stack2.empty())
        	stack2.pop();
        else 
        {
			while(!stack1.empty())
			{
				stack2.push(stack1.pop());
			}
			stack2.pop();
		}
    }

    // Get the front element.
    public int peek() {
    	int num=0;
    	if(!stack2.empty())
        	num= stack2.peek();
        else 
        {
			while(!stack1.empty())
			{
				stack2.push(stack1.pop());
			}
			num= stack2.peek();
		}
        return num;
    }

    // Return whether the queue is empty.
    public boolean empty() {
        if(stack1.empty()&&stack2.empty())
        	return true;
        return false;
    }
}

leetcode_Implement Queue using Stacks

标签：implement queue usin   stack   leetcode   

原文地址：http://blog.csdn.net/mnmlist/article/details/47333803