[92. Reverse Linked List II] #leetcode

92_1023.py

class Solution:
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        # corner case
        if m == n:
            return head
        
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy
        
        # go to mth node
        for i in range(m-1):
            pre = pre.next
            
        # reverse
        reverse = None
        cur = pre.next
        
        for i in range(n-m+1):
            nxt = cur.next
            cur.next = reverse
            reverse = cur
            cur = nxt
            
        pre.next.next = cur
        pre.next = reverse
        
        return dummy.next