算法：链表的逆转

LeetCode OJ Problem: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.

class Solution {
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
ListNode *p, *pre_p, *r, *head2;
int temp;
pre_p = p = r = head;
if(m == n)
return head;
while(m > 1)
{
pre_p = p;
p = p->next;
m--;
}
while(n > 1)
{
r = r->next;
n--;
}

head2 = r->next;
r->next = NULL;

if( pre_p != p)//第一个要逆转的结点不为头结点
{
pre_p->next = NULL;
reverseList(p, r);

pre_p->next = p;
r->next = head2;
}
else
{
reverseList(head, r);

r->next = head2;
}

return head;

}

void reverseList(ListNode *&head, ListNode *&end){
if((head == NULL) || (head->next == NULL))
return ;
ListNode *p, *q, *r;
end = p = head;
r = head->next;
head->next = NULL;
while(r != NULL)
{
q = r;
r = r->next;
q->next = p;
p = q;
}

head = p;

}
};