Problem:
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.
Solution:逆转用的头插法
/**
* 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 result(-1);
result.next=head;
ListNode *prev=&result;
for(int i=0;i<m-1;i++)
prev=prev->next;
ListNode *head2=prev;
prev=head2->next;
ListNode *cur=prev->next;
for(int i=m;i<n;i++)
{
prev->next=cur->next;
cur->next=head2->next;
head2->next=cur; //头插法
cur=prev->next;
}
return result.next;
}
};
原文：http://www.cnblogs.com/xiaoying1245970347/p/4597206.html