92. Reverse Linked List II

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本文介绍了一种在给定范围内反转链表的有效算法。该算法能在一次遍历内完成任务，并保持链表的原有内存地址不变。通过实例展示了如何将1->2->3->4->5->NULL，在m=2且n=4的情况下反转为1->4->3->2->5->NULL。

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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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL || head->next == NULL || n - m < 1)
            return head;

        ListNode dummy(-1);
        dummy.next = head;

        ListNode *start = &dummy, *cur = head, *end = head;
        for (int i = 1; i < m; ++i) {
            start = start->next;
            cur = cur->next;
        }

        for (int j = 0; j < n; ++j)
            end = end->next;

        ListNode *prev = NULL;
        for (int k = m; k <= n; ++k) {
            ListNode *next = cur->next;
            cur->next = prev ? prev : end;
            prev = cur;
            cur = next;
        }

        start->next = prev;
        return dummy.next;
    }
};