92. Reverse Linked List II

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本文介绍了一种在链表中从位置m到n进行就地反转的方法，并提供了一个高效的C++实现方案。

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1.问题描述

Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

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

2.题目分析

题目分析：链表中区间为[m,n]的元素进行反转，因此我们先定位到这个区间（leftpre,right],使用206题的方法进行反转。最后进行两个链表的拼接。

3.c++代码

//我的代码：（beats 100%）
ListNode* reverseBetween(ListNode* head, int m, int n)
{
    if (head == NULL)return head;
    ListNode*dummy = new ListNode(0);
    dummy->next = head;
    ListNode*leftpre = dummy;
    ListNode*right = dummy;
    for (int i = m; i <= n; i++)
        right = right->next;
    for (int j = 1; j < m; j++)
    {
        leftpre = leftpre->next;
        right = right->next;
    }
    //leftpre->next = right->next;
    ListNode*newhead = new ListNode(leftpre->next->val);
    newhead->next = right->next;
    right->next = NULL;
    ListNode*p = leftpre->next->next;
    ListNode*tmp;
    while (p)
    {
        tmp = p->next;
        p->next = newhead;//头插法
        newhead = p;
        p = tmp;
    }
    leftpre->next = newhead;
    return dummy->next;
}