sort list

最新推荐文章于 2022-03-08 09:03:19 发布

原创最新推荐文章于 2022-03-08 09:03:19 发布 · 306 阅读

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本文介绍了一种在O(nlogn)时间内使用常数空间复杂度对链表进行排序的方法。该方法采用归并排序的思想，通过快慢指针找到中间节点将链表一分为二，递归排序左右两部分链表后再合并。文章提供了完整的C++实现代码。

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

//merge sort

/**
 * 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;
        ListNode* fast = head, *slow = head;
        while (fast->next != NULL  && fast->next->next != NULL)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
       // break list       
        fast = slow;
        slow = slow->next;
        fast->next = NULL;

        ListNode* left = sortList(head);
        ListNode* right = sortList(slow);

        return mergeTwoLists(left,right);
    }

    ListNode* mergeTwoLists(ListNode* left, ListNode* right)
    {
        ListNode  dummy(-1);
        for (ListNode* p = &dummy; left != NULL || right != NULL; p = p->next)
        {
            int val1 = left == NULL ? INT_MAX : left->val;
            int val2 = right == NULL ? INT_MAX : right->val;
            if (val1 <= val2)
            {
               p->next = left;
               left = left->next;
            }
            else
            {
                p->next = right;
                right = right->next;
            }
        }
        return dummy.next;
    }


};