Sort List (Medium)

Description

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

Analysis

对单向链表进行排序，时间复杂度为nlogn,其中快排,堆排序,合并排序都可以达到这个复杂度。而快排不太适合单向链表。我们可以用合并排序来实现。

My Solution

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//C++
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution
{
public:
    ListNode *sortList(ListNode *head)
    {
        if(head == NULL || head->next == NULL)
            return head;
        else
        {
            ListNode *p = head,*mid = head, *tmp=head;
            while(tmp->next != NULL && tmp->next->next != NULL)
            {
                tmp = tmp->next->next;
                mid = mid->next;
            }
            tmp = mid;
            mid = mid->next;
            tmp->next = NULL;
            tmp = sortList(p);
            mid = sortList(mid);
            return listmerge(tmp,mid);
        }
    }
    ListNode *listmerge(ListNode *head1, ListNode *head2)
    {
        if(head1 == NULL)
            return head2;
        if(head2 == NULL)
            return head1;
        ListNode *head=NULL , *p ;
        if(head1->val < head2->val)
        {
            head = head1;
            head1 = head1->next;
        }
        else
        {
            head = head2;
            head2 = head2->next;
        }
        p = head;

        while(head1 != NULL && head2 != NULL)
        {
            if(head1->val < head2->val)
            {
                p->next = head1;
                head1 = head1->next;
            }
            else
            {
                p->next = head2;
                head2 = head2->next;
            }
            p = p->next;
        }
        while(head1 != NULL)
        {
            p->next = head1;
            head1 = head1->next;
            p = p->next;
        }
        while(head2 != NULL)
        {
            p->next = head2;
            head2 = head2->next;
            p = p->next;
        }
        p->next = NULL;
        return head;
    }
};