本文主要是介绍Java | Leetcode Java题解之第148题排序链表，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目：
题解：
class Solution {public ListNode sortList(ListNode head) {if (head == null) {return head;}int length = 0;ListNode node = head;while (node != null) {length++;node = node.next;}ListNode dummyHead = new ListNode(0, head);for (int subLength = 1; subLength < length; subLength <<= 1) {ListNode prev = dummyHead, curr = dummyHead.next;while (curr != null) {ListNode head1 = curr;for (int i = 1; i < subLength && curr.next != null; i++) {curr = curr.next;}ListNode head2 = curr.next;curr.next = null;curr = head2;for (int i = 1; i < subLength && curr != null && curr.next != null; i++) {curr = curr.next;}ListNode next = null;if (curr != null) {next = curr.next;curr.next = null;}ListNode merged = merge(head1, head2);prev.next = merged;while (prev.next != null) {prev = prev.next;}curr = next;}}return dummyHead.next;}public ListNode merge(ListNode head1, ListNode head2) {ListNode dummyHead = new ListNode(0);ListNode temp = dummyHead, temp1 = head1, temp2 = head2;while (temp1 != null && temp2 != null) {if (temp1.val <= temp2.val) {temp.next = temp1;temp1 = temp1.next;} else {temp.next = temp2;temp2 = temp2.next;}temp = temp.next;}if (temp1 != null) {temp.next = temp1;} else if (temp2 != null) {temp.next = temp2;}return dummyHead.next;}
}这篇关于Java | Leetcode Java题解之第148题排序链表的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
