LeetCode 92. Reverse Linked List II

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本文介绍了一种在单次遍历中反转链表指定区间的算法实现。通过使用虚拟头节点简化边界条件处理，并通过迭代方式调整节点指针，实现从位置m到n的链表部分的反转。

/**
 * 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 (m == n) {
            return head;
        }
        ListNode dummy(0);
        ListNode *pre = &dummy;
        dummy.next = head;
        for (int i = 1; i < m; ++i) {
            pre = pre->next;
        }
        head = pre->next;
        ListNode *cur = head->next;
        for (int i = m; i < n; ++i) {
            ListNode *temp = cur->next;
            cur->next = pre->next;
            pre->next = cur;
            cur = temp;
        }
        head->next = cur;
        return dummy.next;
    }
};

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.