题解 | 单链表的排序

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 * */
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 == nullptr || head->next == nullptr) {
            return head;
        }
        return mergeSort(head);
    }

    ListNode* mergeSort(ListNode* head) {
        if (head->next == nullptr) {
            return head;
        }

        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* pre = nullptr;
        while (fast != nullptr && fast->next != nullptr) {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = nullptr;
        ListNode* first = mergeSort(head);
        ListNode* second = mergeSort(slow);
        return merge(first, second);
    }

    ListNode* merge(ListNode* first, ListNode* second) {
        auto fake = ListNode(0);
        ListNode* p = &fake;
        while (first != nullptr && second != nullptr) {
            if (first->val <= second->val) {
                p->next = first;
                first = first->next;
            } else {
                p->next = second;
                second = second->next;
            }
            p = p->next;
        }
        if (first != nullptr) {
            p->next = first;
        }
        if (second != nullptr) {
            p->next = second;    
        }
        return fake.next;
    }
};