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.
题意：将链表m到n之间的结点逆置。

思路：链表实现。注意：可以使用辅助。

/**
* 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 *assist = new ListNode(-1), *pre = NULL, *cur = assist, *first = NULL, *last = NULL;
assist->next = head;
for (int i = 0; i <= n; i++){
if (i < m - 1){
cur = cur->next;
}
else if (i == m - 1){
first = cur;
cur = cur->next;
}
else{
if (i == n)
last = cur;
ListNode* tmp = cur->next;
cur->next = pre;
pre = cur;
cur = tmp;
}
}
first->next->next = cur;
first->next = last;
return assist->next;
}
};