题解 | #单链表的排序#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <cstddef>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 the head node
     * @return ListNode类
     */

    ListNode* sortInList(ListNode* head) {
        // write code here
        if(!head||!head->next) return head;
        ListNode* fast = head->next,*slow = head;
        while (fast&&fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode *tmp = slow->next;
        slow->next = NULL;
        ListNode *left = sortInList(head);
        ListNode *right = sortInList(tmp);
        ListNode *h = new ListNode(0);
        ListNode *res = h;
        while (left&&right) {
            if (left->val<right->val) {
                h->next = left;
                left = left->next;
            }else{
                h->next = right;
                right = right->next;
            }
            h = h->next;
        }
        h->next = left ? left : right;
        return res->next;
    }
};

思路复盘：SortInList对链表二分分解，while相当于递归，对链表进行合并