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