【LeetCode with Python】 Reverse Linked List II

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原题页面：https://oj.leetcode.com/problems/reverse-linked-list-ii/
题目类型：
难度评价：★
本文地址：http://blog.csdn.net/nerv3x3/article/details/4223089

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.

反转链表中的一部分。貌似写出来的代码有点长，也许可以优化。

class Solution:    # @param head, a ListNode    # @param m, an integer    # @param n, an integer    # @return a ListNode    def reverseBetween(self, head, m, n):        if m == n:            return head        m_is_head = True if 1 == m else False        if False == m_is_head:            m_pre_node = head            for i in range(0, m - 1 - 1):                m_pre_node = m_pre_node.next            m_node = m_pre_node.next        else:            m_node = head        n_node = m_node        for i in range(0, n - m):            n_node = n_node.next        n_next_node = n_node.next        if False == m_is_head:            m_pre_node.next = n_node        p1_node = m_node        p2_node = p1_node.next        p3_node = p2_node.next        while n_node != p1_node:            p2_node.next = p1_node            p1_node = p2_node            p2_node = p3_node            if None == p2_node:                break            p3_node = p3_node.next        m_node.next = n_next_node        return n_node if m_is_head else head