leetcode——Reverse Linked List II 選擇鏈表中部分節點逆序（AC）

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.

處理這個問題還是挺復雜的，需要考慮很多邊界的測試用例。我總體的思路是先用循環標記m前一個節點和n後邊一個節點，把n後邊的節點首先作為當前逆轉節點的pre，然後循環n-m次完成所選節點部分的逆序，然後將標記的m節點前一個節點指向逆序後部分的頭節點即可。要考慮各種特殊情況，另外考慮即可。code如下：

class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if(head==NULL || m<0 || n<0)
            return head;
        if(head->next == NULL || m==n)
            return head;
        ListNode *head2=NULL,*pre,*cur,*temp=head;
        for(int i=0; inext;
        }
        for(int i=m;inext;
            cur->next = pre;
            pre = cur;
            cur = temp;
        }
        if(m==1)
            return pre;
        head2->next = pre;
        return head;
    }
};