148. Sort List

148. Sort List

Sort a linked list in O(n log n) time using constant space complexity.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4
Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

此题需要时间复杂度为O(n log n)，那么用归并排序。每次都在链表的中间截取，对前后进行排序再拼接。

class Solution {
    public ListNode devide(ListNode head) {
// 剩下两个元素，判断返回
        if (head == null ||  head.next == null) {
            return head;
        }
        if (head.next.next == null) {
            if (head.val > head.next.val) {
                ListNode headNext = head.next;
                headNext.next = head;
                head.next = null;
                head = headNext;
            }
            return head;
        }
// 快慢指针找中点
        ListNode mid = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            mid = mid.next;
            fast = fast.next.next;
        }
        ListNode midNext = mid.next;
        mid.next = null;
// 分段排序
        ListNode sortedFront = devide(head);
        ListNode sortedTail = devide(midNext);
// 合并
        return merge(sortedFront, sortedTail);
    }

    private ListNode merge(ListNode sortedFront, ListNode sortedTail) {
        ListNode begin = new ListNode(-1);
        ListNode cur = begin;
        while (sortedFront != null || sortedTail != null) {
            if (sortedFront != null && sortedTail == null) {
                cur.next = sortedFront;
                sortedFront = sortedFront.next;
                cur = cur.next;
            } else if (sortedFront == null && sortedTail != null) {
                cur.next = sortedTail;
                sortedTail = sortedTail.next;
                cur = cur.next;
            } else {
                if (sortedFront.val <= sortedTail.val) {
                    cur.next = sortedFront;
                    sortedFront = sortedFront.next;
                    cur = cur.next;
                } else {
                    cur.next = sortedTail;
                    sortedTail = sortedTail.next;
                    cur = cur.next;
                }
            }
        }
        return begin.next;
    }

    public ListNode sortList(ListNode head) {
        return devide(head);
    }
}

 

来源：CSDN

作者：粽子包子粿条

链接：https://blog.csdn.net/lianggx3/article/details/104463590