Leetcode:Sort List 对单链表归并排序

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Sort a linked list in O(n log n) time using constant space complexity.

看到O(n log n)的排序算法，适合单链表的首先想到的就是归并排序

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode * findListMid(ListNode * head)    {        if(head == NULL)return NULL;        ListNode * fast = head;        ListNode * slow = head;        while(fast->next)        {            if(fast->next->next)            {                fast = fast->next->next;                slow = slow->next;            }            else                return slow;        }        return slow;    }    ListNode * merge(ListNode * list1,ListNode * list2)    {        if(list1 == NULL)return list2;        if(list2 == NULL)return list1;        ListNode * head = NULL;        if(list1->val >= list2->val)        {            head = list2;            list2 = list2->next;        }        else            if(list1->val < list2->val)            {                head = list1;                list1 = list1->next;            }            ListNode * tail = head;        while(list1 && list2)        {            if(list1->val >= list2->val)            {                tail->next = list2;                tail = list2;                list2 = list2->next;            }              else                if(list1->val < list2->val)                {                    tail->next = list1;                    tail = list1;                    list1 = list1->next;                }          }        if(list1)            tail->next = list1;        if(list2)            tail->next = list2;        return head;    }    ListNode *sortList(ListNode *head) {        if(head == NULL)return NULL;        if(head->next == NULL)return head;        ListNode * mid = findListMid(head);        ListNode * m = mid->next;        mid->next = NULL;        ListNode * list1 = sortList(head);        ListNode * list2 = sortList(m);        ListNode * list = merge(list1,list2);        return list;    }};

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