[leetcode][list][two pointers] 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.

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
//m == 1时head会变
ListNode* reverseBetween(ListNode* head, int m, int n) {
if(NULL == head || NULL == head->next || m == n) return head;
ListNode *p = head;//用于遍历链表
ListNode *pre = NULL;//指向当前链表的前一个
int i = 1;//节点个数计数器
while(i < m && p){//找到需要翻转的第一个节点
pre = p;
p = p->next;
++i;
}
ListNode *first = reverse(p, n-m+1);//返回翻转后的第一节点
if(pre) pre->next = first;//将翻转后的链表连入原链表
else head = first;//m == 1时pre == NULL
return head;
}
private:
ListNode *reverse(ListNode *head, int len){//翻转从head开始的len个节点
if(NULL == head || NULL == head->next) return head;
ListNode *pFirst = head;//新链表的第一个节点，维护它一是为了将节点连入新链表，二是为了返回新链表
ListNode *p = head->next;//当前节点，用于遍历链表
while(--len && p){
ListNode *pNext = p->next;
p->next = pFirst;
pFirst = p;
p = pNext;
}
head->next = p;//现在head是新链表的最后一个节点，它的next是原链表最后一个节点的next
return pFirst;
}
};