148. 排序链表（高频题）

148. 排序链表

• 解题思路
• 代码

解题思路

这题想到要排序，是链表。用快慢指针，还需要扩展一下，归并排序。
但是要注意，这里的归并排序和 给数组归并排序有一丢丢的不同，因为这个链表不支持随机访问

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        return Sort(head);
    }

    /**
     * 返回拆分后的链表头节点，方便归并排序
     *
     * @param head
     * @return
     */
    public ListNode Sort(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode tempNext = slow.next;
        slow.next = null;

        //找到两个新的起点,二路归并
        ListNode leftNode = Sort(head);
        ListNode rightNode = Sort(tempNext);

        return merge(leftNode, rightNode);
    }

    public ListNode merge(ListNode leftNode, ListNode rightNode) {
        ListNode prev = new ListNode(-1);//哨兵节点
        ListNode tmp = prev;
        while (leftNode != null && rightNode != null) {

            if (leftNode.val <= rightNode.val) {
                prev.next = leftNode;
                leftNode = leftNode.next;
            } else {
                prev.next = rightNode;
                rightNode = rightNode.next;
            }
            prev = prev.next;
        }
        prev.next = (leftNode == null) ? rightNode : leftNode;

        return tmp.next;
    }
}