Reverse Linked List II

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本文介绍了一种在不使用额外内存的情况下，一次遍历反转链表中从位置m到n的部分节点的方法。

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Reverse Linked List II

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.

 ListNode *reverseBetween(ListNode *head, int m, int n) 
    {
        ListNode dummy(-1);
        dummy.next = head;
        ListNode *prev = &dummy;
        for (int i = 0; i < m-1; ++i)
            prev = prev->next;
        ListNode* const head2 = prev;
        prev = head2->next;
        ListNode *cur = prev->next;
        for (int i = m; i < n; ++i) 
        {
            prev->next = cur->next;
            cur->next = head2->next;
            head2->next = cur; // 头插法
            cur = prev->next;
        }
        return dummy.next;
    }