LeetCode--Python解析【Reverse Linked List】(206)

题目：

方法一：

考察链表的操作和遍历

新建一个链表

创建一个头节点，头节点不用来储存数据，只用指针域

然后每次将原始链表的元素插入新链表的头节点后的位置

最后返回主链表

结束

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        
        new_head = ListNode(None)
        while head: 
            temp = new_head.next
            new_head.next = ListNode(head.val)
            new_head.next.next = temp
            head = head.next
        return new_head.next
            

方法二：

第二种方法首先创建一个新的list

然后将原始链表中的value全部遍历进新的list中

然后每次取list中的最后一个值，放进链表中

更新链表后返回

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        node = head
        temp = []
        while node:
            temp.append(node.val)
            node = node.next
        node = head
        while temp:
            node.val = temp.pop()
            node = node.next
        return head