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[7/21] 92. Reverse Linked List II

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Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.

 

Example 1:

Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]

Example 2:

Input: head = [5], left = 1, right = 1
Output: [5]

 

Constraints:

  • The number of nodes in the list is n.
  • 1 <= n <= 500
  • -500 <= Node.val <= 500
  • 1 <= left <= right <= n

 

Follow up: Could you do it in one pass?
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학부유학생님의 댓글

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Runtime: 47 ms, faster than 55.30% of Python3 online submissions for Reverse Linked List II.
Memory Usage: 14 MB, less than 87.01% of Python3 online submissions for Reverse Linked List II.
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
        if left == right: return head
        
        i = 1
        sub_prev = ListNode(0,head)
        sub_tail = sub_head = head
        
        while i < left:
            sub_prev = sub_prev.next
            sub_tail = sub_tail.next
            sub_head = sub_head.next
            i+=1
        while i < right:
            sub_tail = sub_tail.next
            i+=1
        sub_next = sub_tail.next
        
        # reverse sub linked list
        
        prev, curr, nxt = None, sub_head, sub_head.next
        
        while curr != sub_next:
            curr.next = prev
            prev = curr
            curr = nxt
            nxt = nxt.next if curr else None
        
        sub_prev.next = sub_tail
        sub_head.next = sub_next
        
        return head if left > 1 else sub_tail 
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