LeetCode[Linked List]: Reverse Linked List II

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#LeetCode #Linked List #Reverse #算法 #链表

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本文介绍了一种在链表中反转指定区间的节点，并在原地进行操作且仅需一次遍历的方法。通过使用虚拟头结点技术，使得处理头结点变得简单。具体步骤包括初始化快慢指针，快指针前移n步，慢指针前移m步，然后依次将慢指针和快指针之间的节点插入快指针之后。

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.

思路很简单：设置两个指针：fast往前移动n步，low往前移动m步，然后依次将low和fast之间的节点挨个插到fast的后面即可。注意：为去除头结点的特殊性，需要用到虚拟头结点技术。

ListNode *reverseBetween(ListNode *head, int m, int n) {
    ListNode *dummyHead = new ListNode(0);
    dummyHead->next = head;

    ListNode *fast = head;
    for (int i = 0; i < n - 1; ++i)
        fast = fast->next;

    ListNode *low  = dummyHead;
    for (int i = 0; i < m - 1; ++i)
        low  = low ->next;

    for (int i = 0; i < n - m; ++i) {
        ListNode *curr = low->next;
        low ->next = low ->next->next;
        curr->next = fast->next;
        fast->next = curr;
    }

    head = dummyHead->next;
    delete dummyHead;
    return head;
}