[leetcode] 92. Reverse Linked List II @ python

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原题

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

解法1

先将链表转化为列表, 然后将列表的m至n这一段反转, 之后再转化为链表.
Time: 2* O(n)
Space: O(1)

代码

# 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
        """
        l = []
        while head:
            l.append(head.val)
            head = head.next
        l[m-1:n] = l[m-1:n][::-1]
        dummy = p = ListNode(0)
        
        for val in l:
            p.next = ListNode(val)
            p = p.next
        return dummy.next

解法2

双指针法, 定义pre, cur两个指针, 将这两个指针向前移动m-1次, 然后反转n-m次.
Time: O(n)
Space: O(1)

代码

class Solution:
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        dummy = ListNode(0)
        dummy.next = head
        
        pre, cur = dummy, head
        for i in range(m-1):
            pre = pre.next
            cur = cur.next
        for j in range(n-m):
            temp = cur.next
            cur.next = temp.next
            temp.next = pre.next
            pre.next = temp
            
        return dummy.next