92. Reverse Linked List II

题目：从m到n进行链表旋转

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.

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/**
* 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;

if(n < m || m < 1 || n < 1) return head;

ListNode* tmpN = head;
ListNode* tmpNH = NULL;
ListNode* tmpV = NULL;
ListNode* tmpE = NULL;
bool flag = true;
bool flag1 =true;

int numNode = 0;
while(tmpN != NULL)
{
++numNode;

if(numNode >= m && numNode <= n)
{
flag = false;
ListNode* tmpL = tmpN;

if(flag1)
{
tmpE = tmpN;
flag1 = false;
}

tmpN = tmpN->next;

tmpL->next = tmpV;

tmpV = tmpL;

}else
{
if(flag)
{
tmpNH = tmpN;
}else
{
break;
}

tmpN = tmpN->next;
}

}

if(tmpNH)
tmpNH->next = tmpV;
else
head = tmpV;

if(tmpE) tmpE->next = tmpN;

return head;
}
};