LeetCode—单链表翻转

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 == NULL)            return NULL;        ListNode * pBeginNode = NULL ;        ListNode * pEndNode = NULL;        ListNode * pmNode = head;        ListNode * ptempNode = NULL;        ListNode * pnNode = NULL;                for(int i = 0; i < (m-1); i++) //<find the node befor mth node        {            pBeginNode = pmNode;            pmNode = pmNode->next;        }        ptempNode = pmNode;                ListNode *pPre = pmNode;        pmNode = pmNode->next;        for(int i = 0; i < (n-m); i++) //<exchange the node         {            ListNode *pNext = pmNode->next;            pmNode->next = pPre;            pPre = pmNode;            pmNode = pNext;        }        ptempNode->next = pmNode;                if(pBeginNode)        {            pBeginNode->next = pPre;        }        else        {            head = pPre;        }        return head;    }};