【LeetCode】88. Sort List

题目描述(Medium)

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

题目链接

https://leetcode.com/problems/sort-list/description/

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

算法分析

归并排序，使用快慢指针找到中间节点，并将前后半段断开，再合并两个链表。

提交代码：

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/**
 * 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 && fast->next->next)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
        
        fast = slow;
        slow = slow->next;
        fast->next = NULL;
        
        ListNode* l1 = sortList(head);
        ListNode* l2 = sortList(slow);
        
        return mergeTwoLists(l1, l2);
    }
    
    
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(-1);
        for (ListNode* p = dummy; l1 != NULL || l2 != NULL; p = p->next)
        {
            int val1 = l1 != NULL ? l1->val : INT_MAX;
            int val2 = l2 != NULL ? l2->val : INT_MAX;
            
            if (val1 <= val2)
            {
                p->next = l1;
                l1 = l1->next;
            }
            else
            {
                p->next = l2;
                l2 = l2->next;
            }
        }
        
        return dummy->next;
        
    }
    
};