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-CSDN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))))));}
}