[LeetCode]题解（python）：092 Reverse Linked List II

题目来源

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https://leetcode.com/problems/reverse-linked-list-ii/

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->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

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

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题意分析

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Input:一个链表

Output:翻转后的链表

Conditions:给定翻转的起始位置和终止位置，翻转链表

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题目思路

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关键是理清变换的思路，注意在链表的操作中，经常设置一个空的头节点。

代码思路如图（注意哪些在变化，至于旋转了，程序员你懂的）

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AC代码（Python）

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

class Solution(object):
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        if head == None or head.next == None:
            return head
        newHead = ListNode(0)
        newHead.next = head
        head1 = newHead
        for i in range(m - 1):
            head1 = head1.next
        p = head1.next
        print p.val
        for i in range(n - m):
            tmp = head1.next
            head1.next = p.next
            p.next = p.next.next
            head1.next.next = tmp
            print p.val, head1.val
        return newHead.next