翻转链表

LintCode 35 题

给出一个链表：1->2->3->null，这个翻转后的链表为：3->2->1->null

使用 python 语言实现：

#coding:utf-8

class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next


class Solution:
    """
    采用循环的方法反转链表
    @param: head: n
    @return: The new head of reversed linked list.
    """
    def reverse(self, head):
        if head is None or head.next is None:  
            return head
        cur = head.next
        head.next = None
        while cur:
            cur_next = cur.next
            cur.next = head
            head = cur
            cur = cur_next
        return head # new head

    """
    采用递归的方法反转链表
    时间复杂度 O(n)
    head 为原链表的头结点，new_head 为反转后链表的头结点  
    """
    def reverse_recursive(self, head):
        if head is None or head.next is None:  
            return head
        else:
            new_head = self.reverse_recursive(head.next)
            head.next.next = head
            head.next = None
            return new_head

# 测试代码1，建立链表：1->2->3->None
head = ListNode(1)               
p1 = ListNode(2)                
p2 = ListNode(3)  
head.next = p1 
p1.next = p2 

# 输出链表：3->2->1->None
s1 = Solution().reverse(head)            
while s1:  
    print s1.val  
    s1 = s1.next

# 测试代码2，建立链表：1->2->3->None
head = ListNode(1)               
p1 = ListNode(2)                
p2 = ListNode(3)  
head.next = p1 
p1.next = p2 

# 输出链表：3->2->1->None
s2 = Solution().reverse_recursive(head)
while s2:  
    print s2.val  
    s2 = s2.next

图解递归方法：