# Sort List

{% embed url="<https://www.lintcode.com/problem/98/>" %}

## Solution 1 (Java)

```java
/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param head: The head of linked list.
     * @return: You should return the head of the sorted linked list, using constant space complexity.
     */
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode mid = findMid(head);
        ListNode right = sortList(mid.next);
        mid.next = null;
        ListNode left = sortList(head);
        return merge(left, right);
    }

    private ListNode findMid(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 left, ListNode right) {
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;

        while (left != null && right != null) {
            if (left.val <= right.val) {
                head.next = left;
                left = left.next;
            } else {
                head.next = right;
                right = right.next;
            }
            head = head.next;
        }
        if (left != null) head.next = left;
        if (right != null) head.next = right;
        return dummy.next;
    }
}
```

### Notes

* This is the **merge sort** version of Sort List.
* Time complexity: `O(nlogn)`

## Solution 2 (Java)

```java
/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param head: The head of linked list.
     * @return: You should return the head of the sorted linked list, using constant space complexity.
     */
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)
            return head;

        ListNode mid = findMid(head);
        ListNode leftDummy = new ListNode(0), leftTail = leftDummy;
        ListNode midDummy = new ListNode(0), midTail = midDummy;
        ListNode rightDummy = new ListNode(0), rightTail = rightDummy;

        while (head != null) {
            if (head.val < mid.val) {
                leftTail.next = head;
                leftTail = head;
            } else if (head.val > mid.val) {
                rightTail.next = head;
                rightTail = head;
            } else {
                midTail.next = head;
                midTail = head;
            }
            head = head.next;
        }

        leftTail.next = null;
        midTail.next = null;
        rightTail.next = null;

        ListNode left = sortList(leftDummy.next);
        ListNode right = sortList(rightDummy.next);
        return concat(left, midDummy.next, right);
    }

    private ListNode findMid(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 concat(ListNode left, ListNode mid, ListNode right) {
        ListNode dummy = new ListNode(0), tail = dummy;
        tail.next = left; tail = moveTail(tail);
        tail.next = mid; tail = moveTail(tail);
        tail.next = right; tail = moveTail(tail);
        return dummy.next;
    }

    private ListNode moveTail(ListNode tail) {
        if (tail == null) return null;
        while (tail.next != null) {
            tail = tail.next;
        }
        return tail;
    }
}
```

### Notes

* **Quick sort**


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