LeetCode OJ

题目：

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.

解题思路：

先找到需要翻转的起始节点，然后，翻转其后的n - m 个节点。 注意处理翻转的起始节点为head的情况。

代码：

/**
 * 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) return NULL;
        
        ListNode *pre_node = NULL, *mid_first = head;
        for (int i = 1; i < m; i++) {
            pre_node = mid_first;
            mid_first = mid_first->next;
        }
        ListNode *last = mid_first->next, *pre = mid_first;
        for (int i = m; i < n; i++) {
            ListNode *tmp = last->next;
            last->next = pre;
            pre = last;
            last = tmp;
        }
        if (pre_node == NULL) {
            head = pre;
            mid_first->next = last;
        } else {
            pre_node->next = pre;
            mid_first->next = last;
        }
        return head;
    }
};