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.

思路：反转部分链表。最关键的就是四个点m-1，m，n，n+1，只要把这四个点弄明白后，然后改变m到n间的指针方向，就OK了。首先找出m-1的点，然后找出m点，经循环后得到n和n+1的点。

/**
* 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 *pPre=NULL;//找出m-1点的位置
ListNode *pList=head;
for(int i=1;i<m;i++)
{
pPre=pList;
pList=pList->next;
}
ListNode *pLeft=pList; //m点位置
ListNode *pRight=pList;//找出n点的位置
ListNode *pTail;//找出n+1点的位置
for(int i=m;i<=n;i++)
{
pTail=pList->next;
pList->next=pRight;
pRight=pList;
pList=pTail;
}
if(pPre==NULL)
{
head=pRight;
}
else
{
pPre->next=pRight;
}
pLeft->next=pTail;
return head;
}
};