反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例:

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

解题思路

由于这个问题没有给出限定条件，所以我们必须要在代码中排除一些可能。

if head == None or head.next == None or m >= n or m < 0 or n < 0:
    return head

另外对于m,n超出listNode长度的情况，返回head。而对于m小于listNode长度，而n大于listNode长度，这种情况，我们只反转[m, len(listNode)]部分。

代码写起来比较容易，首先把pre和cur移动到指定位置，然后就是Leetcode 206:反转链表（最详细解决方案！！！）（t1和t2记录pre和cur反转前的位置）

t1   t2       pre  cur  lat
1 -> 2 <- 3 <- 4    5 -> null

接着t1.next=pre和t2.next=cur

t1   t2       pre  cur  lat
     ----------------
     |              |
1    2 <- 3 <- 4    5 -> null
|              |
----------------

class Solution:
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        if head == None or head.next == None or m >= n or m < 0 or n < 0:
            return head

        pre = None
        cur = head
        i = 1
        while i < m and cur != None:
            pre = cur
            cur = cur.next
            i += 1

        t1 = pre
        t2 = cur

        while i <= n and cur != None:
            lat = cur.next
            cur.next = pre
            pre = cur 
            cur = lat
            i += 1

        if m == 1:
            t2.next = cur
            return pre

        t1.next = pre
        t2.next = cur 
        return head

当然我们这里有一个更好的做法，通过构建一个head listNode，这样我们就不用单独讨论m==1的情况

class Solution:
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        if head == None or head.next == None or m >= n or m < 0 or n < 0:
            return head

        h = ListNode(-1)
        h.next = head
        pre = h
        cur = head
        i = 1
        while i < m and cur != None:
            pre = cur
            cur = cur.next
            i += 1

        t1 = pre
        t2 = cur

        while i <= n and cur != None:
            lat = cur.next
            cur.next = pre
            pre = cur 
            cur = lat
            i += 1

        t1.next = pre
        t2.next = cur
        return h.next

我将该问题的其他语言版本添加到了我的GitHub Leetcode

如有问题，希望大家指出！！！