leetcode148. 排序链表

最新推荐文章于 2024-04-03 15:30:21 发布

原创最新推荐文章于 2024-04-03 15:30:21 发布 · 255 阅读

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文章标签：

#链表 #数据结构 #算法

该篇博客详细介绍了如何使用快慢指针对链表进行分治排序，并通过示例展示了具体步骤。首先，利用快慢指针找到链表中间节点，然后将链表分为两部分分别进行排序。接着，通过`mergeTwoLists`函数将两个已排序的链表合并，最终得到升序排列的链表。这种方法巧妙地结合了分治策略和递归思想。

给你链表的头结点 head ，请将其按升序排列并返回排序后的链表。

示例 1：

输入：head = [4,2,1,3]
输出：[1,2,3,4]

示例 2：

输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]

示例 3：

输入：head = []
输出：[]

思路：使用快慢指针，截断节点

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

        ListNode slow = head;
        ListNode quick = head;

        while (quick.next != null && quick.next.next != null) {
            slow = slow.next;
            quick = quick.next.next;
        }

        ListNode next = slow.next;
        slow.next = null;

        ListNode l1 = sortList(head);
        ListNode l2 =sortList(next);
        ListNode listNode = mergeTwoLists(l1, l2);

        return listNode;

    }

    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode head = new ListNode(1);
        ListNode cur = head;
        ListNode cur1 = list1;
        ListNode cur2 = list2;
        while (cur1 != null && cur2 != null) {
            if (cur1.val > cur2.val) {
                cur.next = new ListNode(cur2.val);
                cur = cur.next;
                cur2 = cur2.next;
            } else if (cur1.val < cur2.val) {
                cur.next = new ListNode(cur1.val);
                cur = cur.next;
                cur1 = cur1.next;
            } else {
                cur.next = new ListNode(cur1.val);
                cur.next.next = new ListNode(cur2.val);
                cur = cur.next.next;
                cur1 = cur1.next;
                cur2 = cur2.next;
            }
        }
        while (cur1 != null) {
            cur.next = cur1;
            cur = cur.next;
            cur1 = cur1.next;
        }

        while (cur2 != null) {
            cur.next = cur2;
            cur = cur.next;
            cur2 = cur2.next;
        }
        if (head.next != null) {
            head = head.next;
        } else {
            head = null;
        }

        return head;
    }
}