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.

/**
 * 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) {
        if(!head){
            return head;
        }
        ListNode* newHead = new ListNode(-1);
        newHead->next = head;
        
        
        //因为m 可能为1,n为2,此时需要三个结点才能完成转换，
        //所以pNode设为newHead，
        //这样newHead结点,第一个，第二个结点就可以完成转换
        ListNode* pNode = newHead;
        
        for(int i = 1; i < m ; i++){
            pNode = pNode->next;
        }
        
        ListNode* pm = pNode->next;;
        
        for(int i = m ; i < n; i++){
            ListNode* n = pm->next;
            pm->next = n->next;
            n->next = pNode->next;
            pNode->next = n;
        }
        return newHead->next;
    }
};