leetcode--Sort List

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

Method 1: This code can be accepted by leetcode online judge. The code is the implementation of  the meger sort idea. 

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        ListNode newHead = null;
        int length = 0;
        if(head != null){
            newHead = head;
            ListNode nextNode = head;
            while(nextNode != null){
                ++length;
                nextNode = nextNode.next;
            }
            if(length > 1){
                int rank = 1;
                while(rank < length){
                    ListNode temp = sortListHelper(newHead, rank);
                    newHead = temp;
                    rank *= 2;
                }
            }
        }
        return newHead;
    }    

    public static ListNode sortListHelper(ListNode head, int n){
        ListNode start = head;
        PairNodes sortedList = null;
        while(start != null){
            ListNode start1 = start;
            PairNodes splits = splitNodes(start1, n);
            ListNode firstNode = start;
            ListNode secondNode = null;
            if(splits.head != null)
                secondNode = splits.head;
            PairNodes partialSorted = mergeNodes(firstNode, secondNode);
            if(sortedList == null)
                sortedList = partialSorted;
            else{
                sortedList.tail.next = partialSorted.head;
                sortedList.tail = partialSorted.tail;
            }
            start = splits.tail;
        }
        return sortedList.head;
    }

    public static PairNodes splitNodes(ListNode start, int n){
        PairNodes splits = new PairNodes();
        splits.head = null;
        splits.tail = null;
        int i = 1;
        while(start != null && i < 2*n){
            if(i == n){
                splits.head = start.next;
                start.next = null;
                start = splits.head;
            }
            else
                start = start.next;
            ++i;
        }
        if(start != null){
            splits.tail = start.next;
            start.next = null;
        }
        return splits;
    }

    public static PairNodes mergeNodes(ListNode firstNode, ListNode secondNode){
        PairNodes result = new PairNodes();

        if(firstNode != null || secondNode != null){
            if((firstNode == null && secondNode != null) ||(firstNode != null && secondNode == null)){
                ListNode start = (firstNode == null) ? secondNode : firstNode;
                result.head = start;
                while(start.next != null)
                    start = start.next;
                result.tail = start;            
            }
            else{
                ListNode mini = new ListNode(0);
                ListNode current = mini;
                while(firstNode != null && secondNode != null){
                    if(firstNode.val > secondNode.val){
                        ListNode temp = firstNode;
                        firstNode = secondNode;
                        secondNode = temp;
                    }
                    current.next = firstNode;
                    current = current.next;
                    if(firstNode.next == null)
                        break;
                    else
                        firstNode = firstNode.next;
                }

                firstNode.next = secondNode;
                while(firstNode.next != null)
                    firstNode = firstNode.next;

                result.head = mini.next;
                result.tail = firstNode;
            }
        }
        return result;
    }
}

class PairNodes{
    ListNode head;
    ListNode tail;
    PairNodes(){};
}

　　

 

Method 2: 

The following code is correct and satifies all requirement. But it can not be accepted by leetcode online judge because the recursive calling incurs a java.lang.StackOverflowError.

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
       ListNode newHead = null;
        int length = 0;
        if(head != null){
            newHead = head;
            ListNode nextNode = head;
            while(nextNode != null){
                ++length;
                nextNode = nextNode.next;
            }
            if(length > 1){
                int rank = 1;
                while(rank < length){
                    ListNode temp = sortListHelper(newHead, rank, length);
                    newHead = temp;
                    rank *= 2;
                }
            }
        }
        return newHead;
    }

    public ListNode sortListHelper(ListNode head, int rank, int length){
        ListNode result = head; 
        ListNode firstNode = head;
        ListNode secondNode = head;
        ListNode thirdNode = null;
        ListNode newHead = head;
        if(rank < length){
            int i = 1;
            while(newHead != null && i < 2 * rank){
                if(i == rank){
                    secondNode = newHead.next;
                    newHead.next = null;
                    newHead = secondNode;
                }
                else
                    newHead = newHead.next;
                ++i;
            }
            if(newHead != null){
                thirdNode = newHead.next;
                newHead.next = null;
            }

            if(firstNode.val > secondNode.val){
                ListNode temp = firstNode;
                firstNode = secondNode;
                secondNode = temp;
            }
          
            ListNode mini = new ListNode(0);
            ListNode current = mini;
            while(firstNode != null && secondNode != null){
                if(firstNode.val > secondNode.val){
                    ListNode temp = firstNode;
                    firstNode = secondNode;
                    secondNode = temp;
                }
                current.next = firstNode;
                current = current.next;
                if(firstNode.next == null)
                    break;
                else
                    firstNode = firstNode.next;
            }

            if(secondNode != null)
                firstNode.next = secondNode;
            while(firstNode.next !=null)
                firstNode = firstNode.next;
            
            result = mini.next;   
            length -= i;
            if(thirdNode != null)
                firstNode.next = sortListHelper(thirdNode, rank, length);
            else
                firstNode.next = null;
        }
        return result; 
    }
}