【leetcode】Reverse Linked List II

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.

* Definition for singly-linked list.

* struct ListNode {

* int val;

* ListNode *next;

* ListNode(int x) : val(x), next(NULL) {}

* };

*/

思路：

找到n和m两个点，然后再一对对地交换其val.

class Solution {
public:

ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode *p,*q,*re,*ne;
p=head;
for(int i=1;i<m;i++) //注意从1开始，因为之前已经指向了head，少了一个点。
{
p=p->next;
}
for(int i=m;i<n;i++)
{
q=p;
for(int j=i;j<n;j++)
q=q->next;
swap(p->val,q->val);
n--;
p=p->next;
}
return head;
}
};