LeetCode092. Reverse Linked List II

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

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这个题我用了四个指针记录位置：left和right分别是翻转链表的两边（外），也就是1和5

然后用p和q记录当前要操作的点和这个点的下一个点，也就是2,3

接着就按照常规思路：2.next=5;3.next=2;4.next=3,然后left.next=4就可以了

# 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
        """
        dummy = ListNode(0)
        dummy.next = head
        left = right = dummy
        length=n-m+1

        while m>1:
            left = left.next
            m -= 1
        while n>-1:
            right = right.next
            n -= 1

        p = left.next
        
        while length>0:
            q=p.next
            p.next=right
            right = p
            p = q
            length -= 1
        
        left.next=right
        
        return dummy.next