[Leetcode]92. Reverse Linked List II

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.

Partial list reversal.

1) When the linked list is empty or a node, return to

2) Get the list length and check the M,n range.

3) The head section retains the linked list before the M-node. Second retains the node between [M,n], including m,n two nodes. Next retains the node after the N node.

4) Use **list to conveniently give null after the end of the first part. First is to facilitate the processing of the second paragraph list.

/** * definition for singly-linked list. * struct listnode {  *     int val; *     struct listnode  *next; * }; */struct listnode* reversebetween (Struct ListNode* head,  int m, int n) {    if  ( head == null | |  head->next == NULL )     {         return head;    }        int  len = 0;    struct ListNode *pLen = head;     for  ( ; pLen; pLen = pLen->next )     {         len++;    }         if  ( m  < 1 | |  n > len )     {         Return head;    }        struct listnode  *first = head;    struct ListNode **list = &head;     int cnt = 1;    for  ( ; cnt  < m; cnt++ )     {        list  = & (*list)->next;        first = first- >next;    }    *list = NULL;         struct listnode *second = null;    struct  ListNode *next = NULL;    for  ( ; cnt <=  n; cnt++ )     {        next = first->next ;        first->next = second;         second = first;        first =  next;    }        first = head;     while  ( first != null && first->next !=  NULL )     {        first =  first->next;    }        if  ( head  != NULL )     {        first-> next = second;    }    else    {         head = second;    }         first = head;    while  ( first != NULL &&  first->next != NULL )     {         first = first->next;    }    first->next  = next;        return head;}

[Leetcode]92. Reverse Linked List II