leetcode Sort list-优快云博客

本文链接：https://blog.youkuaiyun.com/pynash123/article/details/84555223

本文介绍了一种使用归并排序思想实现的链表排序算法，该算法可以在O(nlogn)的时间复杂度内完成排序，并且仅使用常数级别的额外空间。通过递归地将链表分为两半，分别排序后再合并的方式进行，适用于单链表结构。

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

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 if (head->next->next == NULL)
        {
            if (head->val > head->next->val)
            {
                head->val ^= head->next->val;
                head->next->val ^= head->val;
                head->val ^= head->next->val;
            }
            return head;
        }
        ListNode * fast = head, * pre = head, *low = head;
        while (fast != NULL && fast->next != NULL)
        {
            pre = low;
            low = low->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        ListNode * node1 = sortList(head);
        ListNode * node2 = sortList(low);
        return merge(node1, node2);
    }
    ListNode* merge(ListNode* l1, ListNode* l2) 
    {
        ListNode *dummy = new ListNode(-1);
        ListNode *cur = dummy;
        while (l1 && l2) 
        {
            if (l1->val < l2->val)
             {
                cur->next = l1;
                l1 = l1->next;
            } 
            else 
            {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        if (l1) 
            cur->next = l1;
        if (l2) 
            cur->next = l2;
        return dummy->next;
    }
};

测试

#include "head.h"
class Solution {
public:
    ListNode *sortList(ListNode *head) 
    {
        if (head == NULL || head->next == NULL)
            return head;
        else if (head->next->next == NULL)
        {
            if (head->val > head->next->val)
            {
                head->val ^= head->next->val;
                head->next->val ^= head->val;
                head->val ^= head->next->val;
            }
            return head;
        }
        ListNode * fast = head, * pre = head, *low = head;
        while (fast != NULL && fast->next != NULL)
        {
            pre = low;
            low = low->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        ListNode * node1 = sortList(head);
        ListNode * node2 = sortList(low);
        return merge(node1, node2);
    }
    ListNode* merge(ListNode* l1, ListNode* l2) 
    {
        ListNode *dummy = new ListNode(-1);
        ListNode *cur = dummy;
        while (l1 && l2) 
        {
            if (l1->val < l2->val)
             {
                cur->next = l1;
                l1 = l1->next;
            } 
            else 
            {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        if (l1) 
            cur->next = l1;
        if (l2) 
            cur->next = l2;
        return dummy->next;
    }
};

int main()
{
    Solution s;
    int A[] = {2,3,657,1,67,6,89};
    vector<int> vec(A, A+7);
    ListNode * list = makelist(vec);
    s.sortList(list);
    printlist(list);
}