[LintCode] Sort List

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.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.
     */
    //思路就是：分治左右两边，然后递归，然后再把左右两边当作sorted listmerge一下
    //用任何递归的想法都是，我写了这个公式的意思就是我已经排序好了。剩下的计算机自己干去
    public ListNode sortList(ListNode head) { 
        //值得注意的地方，第一次没写head.next ＝＝null的情况
        if (head == null || head.next ==null){
            return head;
        }
        ListNode preMid = findMid(head);
        ListNode right = preMid.next;
        preMid.next = null;
        ListNode left = sortList(head);
        right = sortList(right);
        return merge(left, right);
    }
    private ListNode findMid(ListNode head){
        if (head == null){
            return head;
        }
        ListNode fast = head.next;
        ListNode slow = head;
        while (fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
    private ListNode merge(ListNode l1, ListNode l2){
        if (l1 == null){
            return l2;
        }
        if (l2 == null){
            return l1;
        }
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;
        while (l1 != null && l2 != null){
            if (l1.val < l2.val){
                head.next = l1;
                head = head.next;
                l1 = l1.next;
            } else {
                head.next = l2;
                head = head.next;
                l2 = l2.next;
            }
        }
        if (l1 != null){
            head.next = l1;
        }
        if (l2 != null){
            head.next = l2;
        }
        return dummy.next;
    }
}