Leetcode Sort List_sortpickerlist-优快云博客

本文介绍了一种使用归并排序算法对链表进行排序的方法，该方法可以在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。

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

本题使用归并算法，但需要注意的是归并的终结条件。终结条件为head==NULL以及head->next==NULL,因为这两种情况下，程序都不会自动结束。

代码如下：

/**
 * 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 || !head->next)
            return head;
        
        ListNode*slow,*fast,*secondHead;
        slow = fast = head;
        while(fast->next && fast->next->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        secondHead = slow->next;
        slow->next = NULL;
        
        ListNode* left = sortList(head);
        ListNode* right = sortList(secondHead);
        
        return merge(left,right);        
    }
    
    ListNode* merge(ListNode* left,ListNode* right)
    {
        ListNode* head = new ListNode(0);
        ListNode* cur = head;
        
        while(left && right)
        {
            if(left->val < right->val)
            {
                head->next = left;
                left = left->next;
            }
            else
            {
                head->next = right;
                right = right->next;
            }
            head = head->next;
        }
        
        while(left)
        {
            head->next = left;
            left = left->next;
            head = head->next;
        }
        
        while(right)
        {
            head->next = right;
            right = right->next;
            head = head->next;
        }
        
        return cur->next;
    }
};

另一种就是通过长度取半来划分，而不是通过快慢指针，代码如下：

/**
 * 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) {
        int i = 0;
        ListNode* cur = head;
        while(cur)
        {
            cur = cur->next;
            i++;
        }
        return sort(head,i);
    }
    
    ListNode* sort(ListNode* head,int length)
    {
        if(length < 2)
            return head;
        int half = length/2;
        ListNode* cur = head;
        int i=1;
        while(i<half)
        {
            cur = cur->next;
            i++;
        }
        ListNode* right = sort(cur->next,length-half);
        cur->next = NULL;
        ListNode* left = sort(head,half);
        
        return merge(left,right);        
    }
    
    ListNode* merge(ListNode* left,ListNode* right)
    {
        ListNode* head = new ListNode(0);
        ListNode* cur = head;
       
        while(left && right)
        {
            if(left->val < right->val)
            {
                head->next = left;
                left = left->next;
            }
            else
            {
                head->next = right;
                right = right->next;
            }
            head = head->next;
        }
        
        while(left)
        {
            head->next = left;
            left = left->next;
            head = head->next;
        }
        
        while(right)
        {
            head->next = right;
            right = right->next;
            head = head->next;
        }
        
        return cur->next;
    }
};