Leetcode 92. Reverse Linked List II

Description: Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list.

Link: https://leetcode.com/problems/reverse-linked-list-ii/

Examples:

Example:

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

思路: 206. Reverse Linked List 相比，需要找到第m-1, n+1节点，将m-n之间的reverse，head-m-1 重新连接 reversed list，再连n+1到结尾就可以了。当m=1时，m-1就是不合法的节点，所以要特殊考虑这种情况。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        if not head: return head
        if not head.next: return head
        if m == n: return head
        
        if m == 1:
            p = head
            q = p.next
            p.next = None
            length = 1
            while q:
                f = q.next
                q.next = p
                p = q
                q = f
                length += 1
                if length == n:
                    break
            head.next = q
            return p
        else:
            pre = head     
            p = head.next  
            length = 2     # p is the second node, so length start by 2
            while p:
                if length == m:
                    rvs_head = p  # rvs_head is m Node before reverse, afterwards be n Node
                    end = pre     # end is the m-1 Node
                    break
                pre = p
                p = p.next
                length += 1
            q = p.next
            length += 1  # now q is the moving node, and q=p.next make length plus 1
            p.next = None
            while q and length <= n:
                f = q.next
                q.next = p
                p = q
                q = f
                length += 1
            end.next = p    # p is the head of reversed linked list
            rvs_head.next = q # q is the head of list begin by n+1 Node
            return head

显然，这个代码太长了，有明显的冗余，症结在于m-1不合法，那不妨在head前加root节点，即使m=1，m-1可以指向root. 但是下面这个代码，时间和空间都占用更多。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        if not head: return head
        if not head.next: return head
        if m == n: return head
        
        root = ListNode(0)
        root.next = head
        
        pre = root
        p = root.next
        length = 1
        while p:
            if length == m:
                rvs_head = p
                end = pre
                break
            pre = p
            p = p.next
            length += 1
        q = p.next
        length += 1
        p.next = None
        while q and length <= n:
            f = q.next
            q.next = p
            p = q
            q = f
            length += 1
        end.next = p
        rvs_head.next = q
        return root.next

日期: 2020-11-20  今天很充实，很开心