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.

思路：
第一种： 链表头部的前一指针来保存链表头部
第二种： 使用双指针存储开始逆转的前一个的next指针地址
*/

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* reverseBetween(struct ListNode* head, int m, int n){
if(m==n) return head;//n>=m
struct ListNode *h=(struct ListNode*)malloc(sizeof(struct ListNode));
h->next=head;
struct ListNode *cur=head,*insert_r=h,*pre=NULL;
n-=m;
while(--m)
{
insert_r=insert_r->next;//寻找m的前一个节点
}
cur=insert_r->next;
while(n--)
{
pre=cur->next;
cur->next=pre->next;
pre->next=insert_r->next;
insert_r->next=pre;
}
return h->next;
}
struct ListNode* reverseBetween(struct ListNode* head, int m, int n){
if(m==n) return head;//n>=m
struct ListNode **h=&head;
struct ListNode *cur=head,*pre=NULL;
n-=m;
while(m>2)
{
cur=cur->next;
m--;
}
h= m>1 ? &(cur->next) : &head;
cur=*h;

while(n--)
{
pre=cur->next;
cur->next=pre->next;
pre->next=*h;
*h=pre;
}

return head;
}