Reverse Linked List II

问题：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.

这个问题开始题目读错了，以为是互换m，n位置的节点，其实要求逆置m到n区间的节点。

class Solution {
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
if(head == NULL || head->next == NULL || m == n)
return head;
ListNode *pseud_head = new ListNode(-1);
pseud_head->next = head;
int steps = 0;
ListNode *pm, *pre_m, *pn, *pre_n;
ListNode *travel = pseud_head;
while (travel->next != NULL)
{
if(steps + 1 == m)
{
pre_m = travel;
pm = travel->next;
}
if (steps + 1 == n)
{
pre_n = travel;
pn = travel->next;
break;
}
travel = travel->next;
++steps;
}
if(travel != NULL)
{
ListNode *temp = pm;
ListNode *temp_next = pm;
pre_m->next = pn->next;
do
{
temp = temp_next;
temp_next = temp->next;
temp->next = pre_m->next;
pre_m->next = temp;
} while (temp != pn);

}
pm = pseud_head;
pseud_head = pseud_head->next;
delete pm;
return pseud_head;
}
};

这里还是设定了一个表头节点，这样在处理的时候就非常方便，不用特殊处理第一个结点。