【Leetcode】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.

/**
 * 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(-1);
        dummy.next = head;
        ListNode *p = &dummy;
        for (int i = 0; i < m - 1; ++i) {
            p = p->next;
        }
        //p指向第m个结点的前一个结点
        ListNode *t = p->next, *q = t->next;
        //t指向已经reverse的部分的最后一个结点，q指向即将reverse的下一个结点
        for (int i = m; i < n; ++i) {
            t->next = q->next;
            q->next = p->next;
            p->next = q;
            q = t->next;
        }
        return dummy.next;
    }
};

一定要画图分析，小心指针操作。