Sort List

题意：对链表进行排序，要求的时间复杂度为O(n log n)。

解题思路：O(nlogn)的排序有快速排序、归并排序、堆排序。这里使用归并排序，归并排序的基本思想是：

找到链表的中间节点，然后递归对前半部分和后半部分分别进行归并排序，最后对两个以排好序的链表进行Merge。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {   public  ListNode sortList(ListNode head) { if (head == null || head.next == null) return head;//快慢指针找中间节点ListNode fast = head;ListNode slow = head;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}ListNode headB = slow.next;slow.next = null;return  mergeList(sortList(head), sortList(headB));    }    public ListNode mergeList(ListNode headA,ListNode headB) {ListNode dummy = new ListNode(0);ListNode curNode = dummy;//融合两个链表while (headA != null && headB != null) {if (headA.val >= headB.val) {curNode.next = headB;headB = headB.next;}else {curNode.next = headA;headA = headA.next;}curNode = curNode.next;}if (headA != null) {curNode.next = headA;}else if(headB != null){curNode.next = headB;}return dummy.next;}}