链表排序.md

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题目：

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

解析：

先将链表分成两部分，然后在合并这两部分链表

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* merge(ListNode* left, ListNode* right)//合并两部分链表
    {
        ListNode* head = new ListNode(-1);
        ListNode* pre = head;
        while(left != nullptr && right != nullptr)
        {
            if(left->val <= right->val)
            {
                pre->next = left;
                pre = left;
                left = left->next;
            }
            else
            {
                pre->next = right;
                pre = right;
                right = right->next;
            }
        }
        if(left != nullptr)
            pre->next = left;
        if(right != nullptr)
            pre->next = right;
        
        return head->next;
    }
    ListNode* sortList(ListNode* head) {
        if(head == nullptr || head->next == nullptr)
            return head;
        ListNode* pre = nullptr;
        ListNode* slow = head;
        ListNode* fast = head;
        //将链表分成两部分
        while(fast != nullptr && fast->next != nullptr)
        {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = nullptr;//将链表断开，slow为后面一个链表的头
        ListNode* left = sortList(head);
        ListNode* right = sortList(slow);
        
        return merge(left, right);
    }
};