leetcode206 反转链表

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？
方法 1/2

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

class Solution:
    def reverseList(self, head,prev = None):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        #使用递归，执行效率没有下边的办法快
        # curr = head
        # if not head:
        #     return prev
        # curr,head.next = head.next,prev
        # return self.reverseList(curr,head)
        
        if head is None:
            return None
        cur = head
        pre = None
        nxt = cur.next
        while nxt:
            cur.next = pre
            pre = cur
            cur = nxt
            nxt = nxt.next
        cur.next = pre
        head = cur
        return head

方法3

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

class Solution:
    def reverseList(self, head,prev = None):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        res = []
        if not head:
            return None
        while head:
            res.append(head.val)
            head = head.next
        for i in range(int(len(res) / 2)):
            res[i],res[len(res) - i -1] = res[len(res) - i -1],res[i]     
        return res