sort list

Sort a linked list inO(nlogn) time using constant space complexity.

分析

分治法，注意分治法和快排的区别。分治时只要中部分开即可，merge时候会有大小判断。快排需要用pivot把大小区别开来。

用快慢指针找到中部位置。偶数时slow靠后，用prev。不要忘了前半list末尾置为0

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)//排除空和一个元素的情况
            return head;
        ListNode prev = null, slow = head, quick = head;//prev需要初始化
        while(quick != null && quick.next != null){
            prev = slow;//偶数时候slow在偏后，所以用prev来表示slow前位置
            slow = slow.next;
            quick = quick.next.next;
        }
        prev.next = null;//前半list末尾置为0
        ListNode l1 = sortList(head);
        ListNode l2 = sortList(slow);
        ListNode l3 = merge(l1, l2);
        return l3;
    }

    ListNode merge(ListNode l1, ListNode l2){
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;

        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                cur.next = l1;
                l1 = l1.next;
            }else{
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 != null ? l1 : l2;
        return dummy.next;
    }
}