LeetCode 148. 排序链表

148. 排序链表

解题思路

1. 找到链表的中间节点 $head2$ 的前一个节点，并且断开 $head2$ 与其前一个节点的连接，这样就把原链表拆分成两个长度相同的链表
2. 分治、递归调用 sortList，分别排序 $head$ 和 $head2$
3. 之后得到排序好的链表，最后我们合并这两个链表，然后头结点即可

参考代码

class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null) {
            return head;
        }

        ListNode head2 = middleNode(head);
        head = sortList(head);
        head2 = sortList(head2);
        return mergeTwoList(head,head2);
    }

    private ListNode middleNode(ListNode head) {
        ListNode pre = head;
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null) {
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null;
        return slow;
    }

    private ListNode mergeTwoList(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while(list1 != null && list2 != null) {
            if(list1.val < list2.val) {
                cur.next = list1;
                list1 = list1.next;
            } else {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        if(list1 != null) {
            cur.next = list1;
        }
        if(list2 != null) {
            cur.next = list2;
        }
        return dummy.next;
    }
}