反轉從位置 m 到 n 的連結串列。請使用一趟掃描完成反轉。

說明:
1 ≤ m ≤ n ≤ 連結串列長度。

示例:

輸入: 1->2->3->4->5->NULL, m = 2, n = 4
輸出: 1->4->3->2->5->NULL

解題思路

由於這個問題沒有給出限定條件，所以我們必須要在程式碼中排除一些可能。

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

另外對於m,n超出listNode

程式碼寫起來比較容易，首先把pre和cur移動到指定位置，然後就是Leetcode 206:反轉連結串列（最詳細解決方案！！！）（t1和t2記錄pre和cur反轉前的位置）

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

接著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

當然我們這裡有一個更好的做法，通過構建一個head listNode，這樣我們就不用單獨討論m==1的情況

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

如有問題，希望大家指出！！！