Lettcode_252_Implement Stack using Queues
2015-09-20 18:21
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Implement the following operations of a stack using queues.
push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
empty() -- Return whether the stack is empty.
Notes:
You must use only standard operations of a queue -- which means only
and
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
思路:
(1)题意为运用队列来实现栈,主要包括push()、pop()、top()、empty()方法的实现。
(2)该题的思路和“运用栈来实现队列”类似。首先还是得了解栈和队列的特性,这里再复习一遍;栈:先进后出,只能在栈顶添加和删除元素,队列:先进先出,只能在队尾添加元素,从队头移除元素。想用队列实现栈,主要需考虑到如何将先放入的元素先出栈;这里需要用两个队列来实现,其中一个队列在执行操作后为空,另一个队列存放元素。当有元素加入时,如果某一队列不为空,则将元素加入该队列中;当有元素出栈时,此时需从不为空的队列list中将除去队尾元素的其余元素全部加入另一个空的队列list0中,这样,list队列中的队尾元素(对应栈顶元素)就没有被加入到list0中,然后将list置为一个空的队列,即完成出栈操作。其余的操作比较简单,请参见代码。
(3)详情见下方代码。希望本文对你有所帮助。
算法代码实现如下:
Implement the following operations of a stack using queues.
push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
empty() -- Return whether the stack is empty.
Notes:
You must use only standard operations of a queue -- which means only
push to back,
peek/pop from front,
size,
and
is emptyoperations are valid.
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
思路:
(1)题意为运用队列来实现栈,主要包括push()、pop()、top()、empty()方法的实现。
(2)该题的思路和“运用栈来实现队列”类似。首先还是得了解栈和队列的特性,这里再复习一遍;栈:先进后出,只能在栈顶添加和删除元素,队列:先进先出,只能在队尾添加元素,从队头移除元素。想用队列实现栈,主要需考虑到如何将先放入的元素先出栈;这里需要用两个队列来实现,其中一个队列在执行操作后为空,另一个队列存放元素。当有元素加入时,如果某一队列不为空,则将元素加入该队列中;当有元素出栈时,此时需从不为空的队列list中将除去队尾元素的其余元素全部加入另一个空的队列list0中,这样,list队列中的队尾元素(对应栈顶元素)就没有被加入到list0中,然后将list置为一个空的队列,即完成出栈操作。其余的操作比较简单,请参见代码。
(3)详情见下方代码。希望本文对你有所帮助。
算法代码实现如下:
package leetcode; import java.util.LinkedList; import java.util.List; /** * * @author liqqc * */ public class Implement_Stack_using_Queues { // Push element x onto stack. private List<Integer> list = new LinkedList<Integer>(); private List<Integer> list0 = new LinkedList<Integer>(); public void push(int x) { if (list.size() == 0 && list0.size() == 0) { list.add(x); return; } if (list.size() != 0) { list.add(x); return; } if (list0.size() != 0) { list0.add(x); } } // Removes the element on top of the stack. public void pop() { if (list.size() != 0 && list0.size() == 0) { for (int i = 0; i < list.size() - 1; i++) { list0.add(list.get(i)); } list.clear(); return; } if (list.size() == 0 && list0.size() != 0) { for (int i = 0; i < list0.size() - 1; i++) { list.add(list0.get(i)); } list0.clear(); } } // Get the top element. public int top() { if (list.size() != 0 && list0.size() == 0) { return list.get(list.size() - 1); } if (list.size() == 0 && list0.size() != 0) { return list0.get(list0.size() - 1); } return -1; } // Return whether the stack is empty. public boolean empty() { return list0.size() == 0 && list.size() == 0; } }
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