翻转链表中第m个节点到第n个节点的部分

注意事项

m，n满足1 ≤ m ≤ n ≤ 链表长度

下面是一个傻逼解法。。。

/**
 * Definition of singly-linked-list:
 * 
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The head of linked list.
     * @param m: The start position need to reverse.
     * @param n: The end position need to reverse.
     * @return: The new head of partial reversed linked list.
     */
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        // write your code here
        if (head == NULL) {
            return NULL;
        }
        ListNode *mHead;
        ListNode  *mPreHead = NULL;
        
        ListNode *nHead;
        ListNode *nNextHead = NULL;
        
        
            mHead = head;
            nHead = head;
            int i = 1;
        
        while(i < n) {
            if (i < m) {
                if (i == m-1) {
                    mPreHead = mHead;
                }
                mHead = mHead->next;
            }
           
            nHead = nHead->next;
           
             if(i  == n-1) {
                nNextHead = nHead->next;
            }
             i++;
        }
        
        if (mHead == nHead) {
           
        } else if (mHead->next == nHead) {
            /*ListNode *temp = nHead;
            nHead = mHead;
            mHead = temp;*/
            if (mPreHead != NULL) {
                mPreHead->next = nHead;
            } 
         //   if (nNextHead != NULL) {
                mHead->next = nNextHead;
           // }
             nHead->next = mHead;
             
             if (mPreHead == NULL) {
                 return nHead;
             }
             
            return head;
            
        } else {
            ListNode *pre = mHead;
            ListNode *middle = mHead->next;
            ListNode *behind = middle->next;
            if (nNextHead != NULL) {
                pre->next = nNextHead;
            }
         
                middle->next = pre;
                
                pre = middle;
                middle = behind;
                behind = behind->next;
                
          while(behind != nNextHead) {
                middle->next = pre;
                pre = middle;
                middle = behind;
                behind = behind->next;
                if (behind == nNextHead) {
                    middle->next = pre;
                    if (mPreHead != NULL) {
                         mPreHead->next = middle;
                    }
                   if (m==1) {
                       return middle;
                   }
                }
            }
        }
        
         return head;
    }
};