Problem #148: Sort List

Intuition & Strategy Breakdown

Use Merge Sort on linked list. Divide list into two halves using fast/slow pointers, recursively sort both halves, and merge sorted lists.

Python Solution

class Solution:
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next:
            return head
        
        left = head
        mid = self.getMid(head)
        right = mid.next
        mid.next = None
        
        left = self.sortList(left)
        right = self.sortList(right)
        
        return self.merge(left, right)
    
    def getMid(self, head):
        slow, fast = head, head.next
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow
    
    def merge(self, l1, l2):
        dummy = tail = ListNode()
        while l1 and l2:
            if l1.val < l2.val:
                tail.next, l1 = l1, l1.next
            else:
                tail.next, l2 = l2, l2.next
            tail = tail.next
        tail.next = l1 or l2
        return dummy.next

Java Solution

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head;
        
        ListNode mid = getMid(head);
        ListNode right = mid.next;
        mid.next = null;
        
        ListNode left = sortList(head);
        right = sortList(right);
        
        return merge(left, right);
    }
    
    private ListNode getMid(ListNode head) {
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                tail.next = l1;
                l1 = l1.next;
            } else {
                tail.next = l2;
                l2 = l2.next;
            }
            tail = tail.next;
        }
        tail.next = (l1 != null) ? l1 : l2;
        return dummy.next;
    }
}