148:sort list "Sort" of "linked list"

Title Link: https://leetcode.com/problems/sort-list/

/* Test Instructions: Sort the list *//** * Ideas: Merge sort * Divide: Divide the list into two segments: with slow and fast pointers, slow only one step at a time, fast every time * walk two steps. When fast is empty, the slow is located in the middle of the list.  * Merge: Two simple linked list merge */class solution {public:void Show (ListNode *head) {while (head! = NULL) {cout            << head->val << "";        Head = head->next;    } cout << Endl;        } ListNode *merge (ListNode *heada, ListNode *headb) {ListNode *root = new ListNode (0);        ListNode *p = root;                while (Heada && headb) {if (Heada->val <= headb->val) {p->next = Heada;            Heada = heada->next;                } else {p->next = headb;            HEADB = headb->next;        } p = p->next;            } while (Heada) {p->next = Heada;            p = p->next;        Heada = heada->next;            } while (headb) {p->next = headb;           p = p->next; HEADB = headb->next;        } ListNode *res = root->next;        Delete root;    return res;        } ListNode *mergesort (ListNode *head) {///Only one node if (head = = NULL | | head->next = NULL) return head;            ListNode *slow = head;            Slow pointer ListNode *fast = head->next->next;//fast pointer while (FAST! = NULL && Fast->next! = null) {            slow = slow->next;        Fast = fast->next->next;        } ListNode *heada = head;        ListNode *headb = slow->next;      Slow->next = NULL;        Disconnect the list Heada = MergeSort (Heada);        HEADB = MergeSort (HEADB);    Return Merge (Heada, headb);        } listnode* sortlist (listnode* head) {if (head = = NULL | | head->next = = NULL) return head;    Return MergeSort (head); }};

　　

148:sort list "Sort" of "linked list"