92. 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.

1.我的答案

/**
* 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) {
ListNode* r = head, *p = NULL, * q = NULL, *temp = NULL;
if(m == n) return head;
int mm = m;
int i = 1;
while(i < m-1){
r = r -> next;
i++;
}
if(m == 1)
p = r;
else
p = r-> next;
q = p -> next;
while(mm < n){
temp = q->next;
q->next = p;
p = q;
q = temp;
mm++;
}
if(m == 1){
head = p;
temp = r;
}
else{
temp = r -> next;
r->next = p;
}
temp -> next = q;
return head;
}
};

2，别人的答案
ListNode *reverseBetween(ListNode *head, int m, int n) {
if(m==n)return head;
n-=m;
ListNode prehead(0);
prehead.next=head;
ListNode* pre=&prehead;
while(--m)pre=pre->next;
ListNode* pstart=pre->next;
while(n--)
{
ListNode *p=pstart->next;
pstart->next=p->next;
p->next=pre->next;
pre->next=p;
}
return prehead.next;
}