reverse linklist 反转链表的一部分 python

描述：

Reverse a linked list from position m to n. Do it in-place and in one-pass.For example: 

Given 1->2->3->4->5->nullptr, m = 2 and n = 4,

return 1->4->3->2->5->nullptr.Note: 

Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.

思路：对【m, n】区间的数进行头插法插入

代码如下：

# coding=utf-8
"""思路就是针对反转区间的数使用头插法，逻辑比较繁琐"""


class LinkNode(object):  # 定义链表节点
    def __init__(self, val, nextnode=None):
        self.val = val
        self.nextnode = nextnode


class Solution(object):
    def reverse_between(self, linkroot=None, m=0, n=0):
        root = LinkNode(0)  # 处理反转起始位置为1，而增加一个节点
        pre = root
        pre.nextnode = linkroot
        p = linkroot
        t = m - 1
        while t > 0:  # pre 始终记录反转的起始位置的直接前驱
            pre = p
            p = p.nextnode  # 始终保持指向反转的起始位置节点
            t -= 1
        cur = p  # cur记录每次头插法的点
        q = p.nextnode  # q记录反转区间中的点
        for _ in range(m + 1, n + 1):  # 从反转起始的下一个点开始头插法
            if q is not None:
                r = q.nextnode  # r记录q的直接后继节点
                pre.nextnode = q
                q.nextnode = cur
                p.nextnode = r
                cur = q
                q = r
        return root.nextnode
if __name__ == '__main__':
    """1->2->3->4->5->6->7->8->9  to 1->6->5->4->3->2->7->8->9"""
    roota = LinkNode(1)
    pp = roota
    for i in range(2, 10):
        qq = LinkNode(i)
        pp.nextnode = qq
        pp = pp.nextnode
    root = Solution().reverse_between(roota, 2, 6)
    while root is not None:
        print root.val,
        root = root.nextnode