148. 排序链表

https://leetcode.cn/problems/sort-list/description/?envType=study-plan-v2&envId=top-interview-150

一道经典的考察归并排序的题目，采用分治的思想，先拆分成单一的有序部分，再合并这些有序部分成更大的有序部分，对归并排序不了解的可以看看这篇博客：排序算法——归并排序-优快云博客

class Solution {
    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; }
    }
    public ListNode sortList(ListNode head) {
        return divide(head);
    }

    public ListNode divide(ListNode node) {
        if(node == null || node.next == null) {
            return node;
        }
        // 快慢指针找中点
        ListNode slow = node, fast = node;
        ListNode pre = null;
        while(slow != null && fast != null && fast.next != null) {// 快慢指针找到中间节点
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null;// 将链表一分为二
        return merge(divide(node), divide(slow));
    }

    public ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1); // 虚拟头节点
        ListNode curr = dummy;// 当前尾节点
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                curr.next = l1;
                l1 = l1.next;
            } else {
                curr.next = l2;
                l2 = l2.next;
            }
            curr = curr.next;
        }
        //拼接后续链表
        curr.next = (l1 != null) ? l1 : l2;
        return dummy.next;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        System.out.println(solution.sortList(solution.new ListNode(4, solution.new ListNode(2, solution.new ListNode(1, solution.new ListNode(3))))));
    }
}