Leetcode -- 148. Sort List (MergeSort)

148. Sort List(MergeSort)

• Difficulty: Medium

 https://leetcode.com/problems/sort-list/
Sort a linked list in O(n log n) time using constant space complexity.

思路： Use the idea merge sort to recursively half linked list and merge lists
Complexity: O(n log n), Space O(n) for merge

public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)
            return head;
        // step 1. cut the list to two halves
        ListNode prev = null, slow = head, fast = head;//prev is end of 1st half
        while(fast!=null&&fast.next!=null){
            prev = slow;
            slow=slow.next;
            fast=fast.next.next;
        }
        prev.next = null;
        // step 2. sort each half
        ListNode l1 = sortList(head);//if recursion return sth, need to assign it!
        ListNode l2 = sortList(slow);
        // step 3. merge l1 and l2
        return merge(l1, l2);
    }
  
  ListNode merge(ListNode l1, ListNode l2) {
        if(l1==null)return l2;
        if(l2==null)return l1;
        if(l1.val<l2.val){
            l1.next = merge(l1.next,l2);
            return l1;
        }else{
            l2.next = merge(l1,l2.next);
            return l2;
        }
  }
}

Iterative way of merge
ListNode merge(ListNode l1, ListNode l2) {
        ListNode l = new ListNode(0), p = l;
        while(l1!=null&&l2!=null){
            if(l1.val<l2.val){
                p.next = l1;
                l1 = l1.next;
            }else{
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        if(l1!=null)p.next = l1;
        if(l2!=null)p.next = l2;
        return l.next;
  }