# LeetCode – Sort List (Java)

LeetCode - Sort List:

Sort a linked list in O(n log n) time using constant space complexity.

Keys for solving the problem

1. Break the list to two in the middle
2. Recursively sort the two sub lists
3. Merge the two sub lists

This is my accepted answer for the problem.

```package algorithm.sort;   class ListNode { int val; ListNode next;   ListNode(int x) { val = x; next = null; } }   public class SortLinkedList {   // merge sort public static ListNode mergeSortList(ListNode head) {   if (head == null || head.next == null) return head;   // count total number of elements int count = 0; ListNode p = head; while (p != null) { count++; p = p.next; }   // break up to two list int middle = count / 2;   ListNode l = head, r = null; ListNode p2 = head; int countHalf = 0; while (p2 != null) { countHalf++; ListNode next = p2.next;   if (countHalf == middle) { p2.next = null; r = next; } p2 = next; }   // now we have two parts l and r, recursively sort them ListNode h1 = mergeSortList(l); ListNode h2 = mergeSortList(r);   // merge together ListNode merged = merge(h1, h2);   return merged; }   public static ListNode merge(ListNode l, ListNode r) { ListNode p1 = l; ListNode p2 = r;   ListNode fakeHead = new ListNode(100); ListNode pNew = fakeHead;   while (p1 != null || p2 != null) {   if (p1 == null) { pNew.next = new ListNode(p2.val); p2 = p2.next; pNew = pNew.next; } else if (p2 == null) { pNew.next = new ListNode(p1.val); p1 = p1.next; pNew = pNew.next; } else { if (p1.val < p2.val) { // if(fakeHead) pNew.next = new ListNode(p1.val); p1 = p1.next; pNew = pNew.next; } else if (p1.val == p2.val) { pNew.next = new ListNode(p1.val); pNew.next.next = new ListNode(p1.val); pNew = pNew.next.next; p1 = p1.next; p2 = p2.next;   } else { pNew.next = new ListNode(p2.val); p2 = p2.next; pNew = pNew.next; } } }   // printList(fakeHead.next); return fakeHead.next; }   public static void main(String[] args) { ListNode n1 = new ListNode(2); ListNode n2 = new ListNode(3); ListNode n3 = new ListNode(4);   ListNode n4 = new ListNode(3); ListNode n5 = new ListNode(4); ListNode n6 = new ListNode(5);   n1.next = n2; n2.next = n3; n3.next = n4; n4.next = n5; n5.next = n6;   n1 = mergeSortList(n1);   printList(n1); }   public static void printList(ListNode x) { if(x != null){ System.out.print(x.val + " "); while (x.next != null) { System.out.print(x.next.val + " "); x = x.next; } System.out.println(); }   } }```

Output:

2 3 3 4 4 5
Category >> Algorithms
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• Saikiran Narlapuram
• Arnaud Hebert

Works fine if you change int count = 0; with int count = 1;

• henrycan1414

it is the same since p1.val == p2. val

• Neel Sheyal

If the list is empty or only one element in it, then return the list.
Divide the linked list into two parts.
Sort these two parts recursively.
Merge the sorted parts.

• Frank

This line
pNew.next.next = new ListNode(p1.val);

Should be

pNew.next.next = new ListNode(p2.val);

• anchao1987

• tia

A little bit change to merge function works. Create extra nodes is unnecessary

``` public ListNode merge(ListNode l, ListNode r){ ListNode lp = l, rp = r; ListNode newhead = new ListNode(-1); ListNode cur = newhead; while(lp!=null || rp!=null){ if(lp==null){ cur.next = rp; break; }else if(rp==null){ cur.next = lp; break; }else{ if(lp.val <= rp.val){ cur.next = lp; lp = lp.next; cur = cur.next; }else { cur.next = rp; rp = rp.next; cur = cur.next; } } } return newhead.next; } ```

• Raghav

I’m getting stack overflow error for size of length greater than 3.

• Lister

Definitely not correct. You are creating new nodes.

• Wei Qiu

It’s not correct if you consider the space used by call stack.