力扣算法题—148sort-list

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

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

Solution:
　　复杂度为O(nlogn)有快速排序，并归排序，堆排序，对于来列表而言，堆排序最适合了
　　使用快慢指针将链表分为两部分
　　merge01为简洁版

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (head == nullptr || head->next == nullptr)return head;
        ListNode *slow = head, *fast = head, *pre = head;
        while (fast != nullptr && fast->next != nullptr)
        {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = nullptr;//将链表分为两段
        return merge(sortList(head), sortList(slow));
    }
    ListNode* merge01(ListNode* l1, ListNode* l2)
    {
        if (l1 == nullptr || l2 == nullptr)return l1 == nullptr ? l2 : l1;
        if (l1->val < l2->val)
        {
            l1->next = merge(l1->next, l2);
            return l1;
        }
        else
        {
            l2->next = merge(l1, l2->next);
            return l2;
        }
    }
    ListNode* merge02(ListNode* l1, ListNode* l2)
    {
        ListNode* ptr = new ListNode(-1);
        ListNode* p = ptr;
        while (l1 != nullptr && l2 != nullptr)
        {
            if (l1->val < l2->val)
            {
                p->next = l1;
                l1 = l1->next;
            }
            else
            {
                p->next = l2;
                l2 = l2->next;
            }
            p = p->next;
        }
        if (l1 == nullptr)p->next = l2;
        if (l2 == nullptr)p->next = l1;
        return ptr->next;
    }
};