[Leetcode series] flipped Linked List II

Given a linked list and two integers m, n, flip the M node of the linked list to the N node (counting starts from 1 ).

For example, for a given linked list: 1-> 2-> 3-> 4-> 5-> null, and m = 2, n = 4.

Returns 1-> 4-> 3-> 2-> 5-> null.

Assume that M and N meet the constraints: 1 ≤M≤NLength less than or equal to the length of the linked list.

Note: you cannot use extra space, and you can only use the traversal link table once.

 

Algorithm ideas:

The flip process can be divided into three steps:

Invert the pointing relationship of adjacent nodes; that is, 1-> 2 <-3 <-4 5-> null

Point Node 1 to node N (4); that is, 1-> 4-> 3-> 2 5-> null

Point the MTH node (2, cache required) to the n + 1 node (5). That is, 1-> 4-> 3-> 2-> 5-> null

 

/** * 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 || m == n)            return head;        ListNode *p = head;        int count = 1;        while(count < m-1)        {            p = p->next;            count++;        }        ListNode *p1;        if(m == 1)            p1= p;        else        {            p1 = p->next;            count++;        }        ListNode *last = p1;        ListNode *p2 = p1->next;        ListNode *tmp = NULL;        while(p2 && count < n)        {            tmp = p2->next;            p2->next = p1;            p1 = p2;            p2 = tmp;            count++;        }        last->next = tmp;        if(m == 1)            head = p1;        else            p->next = p1;        return head;    }};

 

[Leetcode series] flipped Linked List II