LeetCode OJ - Reverse Linked List II

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;
}
};