148. Sort List

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148. Sort List

Description

Given the head of a linked list, return the list after sorting it in ascending order .

Example 1:

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Input: head = [4,2,1,3]
Output: [1,2,3,4]

Example 2:

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Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]

Example 3:

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Input: head = []
Output: []

Constraints:

  • The number of nodes in the list is in the range [0, 5 * 10^4].
  • -10^5 <= Node.val <= 10^5

Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?

Hints/Notes

Solution

Language: C++

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode* mid = getMid(head);
        ListNode* left = sortList(head);
        ListNode* right = sortList(mid);
        return merge(left, right);
    }

    ListNode* merge(ListNode* list1, ListNode* list2) {
        ListNode dummy(0);
        ListNode* head = &dummy;
        while (list1 && list2) {
            if (list1->val < list2->val) {
                head->next = list1;
                list1 = list1->next;
            } else {
                head->next = list2;
                list2 = list2->next;
            }
            head = head->next;
        }
        if (list1) {
            head->next = list1;
        } else {
            head->next = list2;
        }
        return dummy.next;
    }

    ListNode* getMid(ListNode* head) {
        ListNode* midPrev = nullptr;
        while (head && head->next) {
            midPrev = (midPrev == nullptr) ? head : midPrev->next;
            head = head->next->next;
        }
        ListNode* mid = midPrev->next;
        midPrev->next = nullptr;
        return mid;
    }
};