[LeetCode] 92. Reverse Linked List II_Medium tag: Linked List

最新推荐文章于 2025-05-21 15:08:20 发布

转载最新推荐文章于 2025-05-21 15:08:20 发布 · 85 阅读

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原文链接：http://www.cnblogs.com/Johnsonxiong/p/10799389.html

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

这个题目是在[LeetCode] 206. Reverse Linked List_Easy tag: Linked List做的follow up/提升，也就是要找到相应的点，然后再做m - n 次reverse，最后将linked list排好就可以了。这里我还是用了temp，便于logic上面的理解。也可以利用python的语句特点进行优化。

Code

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def reverseList2(self, head, m, n):
        dummy = ListNode(0)
        dummy.next = head
        head = dummy
        for _ in range(m - 1):
            head = head.next

        pre = head  # find the node before mNode
        mNode = head.next
        curNode = mNode.next

        # reverse from mNode, just like in LeetCode 206
        for _ in range(n - m):
            temp = curNode.next
            curNode.next = mNode
            mNode = curNode
            curNode = temp

        # combine the 前面一节和reverse的一节，以及最后一节
        temp = pre.next
        pre.next = mNode
        temp.next = curNode
        return dummy.next

转载于:https://www.cnblogs.com/Johnsonxiong/p/10799389.html