LeetCode--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：头插法。这是一个翻转链表题目的扩展，在头插法的基础上考虑在一个区间内反转。那么，首先要找到prev指针，然后确定3个指针head2,prev,cur的位置，头插法把cur指针内容插到head2后面，然后更新cur指针。

/**
* 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 dummy(0);
dummy.next=head;
ListNode *prev=&dummy;
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;
cur=prev->next;
}
return dummy.next;
}
};