148. Sort List

Description

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

Example 1:

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

Example 2:

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

Example 3:

Input: head = []
Output: []

Constraints:

• The number of nodes in the list is in the range [0, 5 * 104].
• 105 <= Node.val <= 105

---

Solutions

Solution 1

Intuition

This is a good question

1. you need to find the mid of list, and split it to two lists
2. you need to merge these above lists.

We have a template for merge sort below.

private int[] temp = new int[5_0000];
private void mergeSort(int[] nums, int l, int r) {
    if (l >= r) {
        return;
    }
    int mid = l + r >> 1;
    mergeSort(nums, l, mid);
    mergeSort(nums, mid + 1, r);
    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r) {
        if (nums[i] < nums[j]) {
            temp[k++] = nums[i++];
        } else {
            temp[k++] = nums[j++];
        }
    }
    while (i <= mid) {
        temp[k++] = nums[i++];
    }
    while (j <= r) {
        temp[k++] = nums[j++];
    }
    for (i = l, j = 0; i <= r; i++, j++) {
        nums[i] = temp[j];
    }
}

Code

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode[] halfSplits = split(head);
        ListNode l = sortList(halfSplits[0]);
        ListNode r = sortList(halfSplits[1]);
        return merge(l, r);
    }

    private ListNode[] split(ListNode head) {
        ListNode prev = null, slow = head, fast = head;
        while (fast != null && fast.next != null) {
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode mid = slow;
        prev.next = null;
        return new ListNode[] { head, mid };
    }

    private ListNode merge(ListNode l, ListNode r) {
        ListNode dummy = new ListNode();
        ListNode head = dummy;
        while (l != null && r != null) {
            if (l.val < r.val) {
                head.next = l;
                l = l.next;
            } else {
                head.next = r;
                r = r.next;
            }
            head = head.next;
        }
        if (l != null) {
            head.next = l;
        }
        if (r != null) {
            head.next = r;
        }
        return dummy.next;
    }
}

Complexity

• Time: O(nlogn)
• Space: O(1)

---