# [C + +] leetcode:62 Reverse Linked List II

Source: Internet
Author: User

Title:

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.

Insertion method

Ideas:

Step 1: Find the location of the M node and maintain a node p before the M node so that it is inserted here.

Step 2: Invert until n nodes. The inverse method is to read one node at a time, insert it into the previous position (P->next) of the current linked list M node, and then the original m nodes receive the next node read.

Complexity: O (N), requires several auxiliary pointers, Space O (1)

Attention:

1. The insertion must be inserted in the previous position of the current m node, which is the next position of the P node that was previously maintained.

`Tmp->next = p->next;   It must be p->next here, because each insertion is in the previous position of M, and the first position after p.`
2. Notice the initialization and definition methods of the head node and the first nodes (the difference node and the linked list)

`ListNode dummyhead (0);d ummyhead.next = head; listnode* p = &dummyhead;`

3. The distinction between two data structure nodes consists of a data field that holds the elements and a pointer field that holds the address of the successor node.   Listnode* p; If P is a pointer to the first element of the linear table.P->next represents a pointer to the next node. ListNode p (0); If P is a single node within a linked list.P.next represents a value that holds the address information for another node within the P node.
`* struct ListNode {*     int val; *     ListNode *next; *     listnode (int x): Val (x), Next (NULL) {} *};`
AC Code:

`/** * Definition for singly-linked list. * struct ListNode {*     int val; *     ListNode *next; *     listnode (int x): Val (x), Next (NULL) {} *}; */class Soluti On {public:    listnode *reversebetween (listnode *head, int m, int n) {        //Insert Method        if (head = = NULL) return null;        ListNode dummyhead (0);        Dummyhead.next = head;        listnode* p = &dummyhead;                for (int i = 1; i < m; i++)        {            p = p->next;        }        Head = p->next;  Head is the M node, p is the previous node of M                ///execute insert for        (; m < n; m++)        {            listnode* tmp = head->next;            Head->next = tmp->next;            Tmp->next = p->next;   It must be p->next here, because each insertion is in the previous position of M, and the first position after p.            p->next = tmp;          }                return dummyhead.next;}    ;`

[C + +] leetcode:62 Reverse Linked List II

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