Leetcode linklist Special Topic

This blog will be leetcode on the linklist topic content here, followed by slowly add

One: Leetcode 206 Reverse Linked list two: Leetcode Reverse Linked List II

A: Leetcode 206 Reverse Linked List

Topic:

Reverse a singly linked list.

Code:

Class Solution {public:    listnode* reverselist (listnode* head) {        if (head = = NULL) return null;        ListNode *p = head;        ListNode *pnext = p->next;        P->next = NULL;       The head node needs to point to NULL otherwise time limit while        (pnext! = NULL) {            ListNode *q = pnext->next;            Pnext->next = p;            p = pnext;   Iteration            pnext = q;        }        return p;            }};

II: Leetcode Reverse Linked List II

Topic:

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.

Analysis: This problem is slightly more complicated than a little, but it is also easy to solve

Code:

Class Solution {public:    listnode* Reversebetween (listnode* head, int m, int n) {        if (m = = N) return head;        ListNode *pre = head;        ListNode *p = head;        for (int i = 1; i < m; i++) {    //Find the first M element and its previous element            pre = p;            p = p->next;        }        ListNode *curr = p;        ListNode *pnext = p->next;   Invert the elements that need to be reverse in the middle to take advantage of the idea in reverse linked list for        (int i = m; i < n; i++) {            ListNode *q = pnext->next;< C14/>pnext->next = p;            p = pnext;            Pnext = q;        }        Curr->next = Pnext;        if (pre! = curr) Pre->next = p;      Inequality indicates m!=1  then Pre->next = p, and equality indicates m==1, then head ==p        else head = p;        return head;            };

Leetcode linklist Special Topic