在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序。
示例 1: 输入: 4->2->1->3 输出: 1->2->3->4
示例 2: 输入: -1->5->3->4->0 输出: -1->0->3->4->5
最开始笨拙的想法
代码语言：javascript复制/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode sortList(ListNode head) {
PriorityQueue<Integer> heap =
new PriorityQueue<Integer>(new Comparator<Integer>() {
@Override
public int compare(Integer n1, Integer n2) {
return n1 - n2;
}
});
while (head != null) {
heap.add(head.val);
head = head.next;
}
ListNode pre = new ListNode(0);
ListNode begin = new ListNode(0);
pre = begin;
while (!heap.isEmpty()) {
ListNode node = new ListNode(heap.poll());
begin.next = node;
begin = node;
}
return pre.next;
}
} public static ListNode sortList(ListNode head) {
// 1、递归结束条件
if (head == null || head.next == null) {
return head;
}
// 2、找到链表中间节点并断开链表 & 递归下探
ListNode midNode = middleNode(head);
ListNode rightHead = midNode.next;
midNode.next = null;
ListNode left = sortList(head);
ListNode right = sortList(rightHead);
// 3、当前层业务操作（合并有序链表）
return mergeTwoLists(left, right);
}
// 找到链表中间节点（876. 链表的中间结点）
private static ListNode middleNode(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode slow = head;
ListNode fast = head.next.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
// 合并两个有序链表（21. 合并两个有序链表）
private static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode sentry = new ListNode(-1);
ListNode curr = sentry;
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;
}
curr.next = l1 != null ? l1 : l2;
return sentry.next;
}
