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148. 排序链表
给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。
示例 1：
输入：head = [4,2,1,3] 输出：[1,2,3,4]
示例 2：
输入：head = [-1,5,3,4,0] 输出：[-1,0,3,4,5]
示例 3：
输入：head = [] 输出：[]
提示：
-
链表中节点的数目在范围
[0, 5 * 104]内 -
-105 <= Node.val <= 105
进阶：你可以在 \(O(n*logn)\) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？
思路分析
分支模式，归并排序。
-
一刷
-
二刷
-
三刷
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/**
* Runtime: 4 ms, faster than 62.99% of Java online submissions for Sort List.
*
* Memory Usage: 40.2 MB, less than 89.47% of Java online submissions for Sort List.
*
* @author D瓜哥 · https://www.diguage.com
* @since 2019-10-29 21:05
*/
public ListNode sortList(ListNode head) {
if (Objects.isNull(head) || Objects.isNull(head.next)) {
return head;
}
ListNode prev = null;
ListNode slow = head;
ListNode fast = head;
while (Objects.nonNull(fast) && Objects.nonNull(fast.next)) {
prev = slow;
slow = slow.next;
fast = fast.next.next;
}
prev.next = null;
ListNode firstResult = sortList(head);
ListNode secondResult = sortList(slow);
return merge(firstResult, secondResult);
}
private ListNode merge(ListNode first, ListNode second) {
ListNode result = new ListNode(0);
ListNode tail = result;
while (Objects.nonNull(first) && Objects.nonNull(second)) {
if (first.val < second.val) {
tail.next = first;
first = first.next;
} else {
tail.next = second;
second = second.next;
}
tail = tail.next;
}
if (Objects.isNull(first)) {
tail.next = second;
}
if (Objects.isNull(second)) {
tail.next = first;
}
return result.next;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-04-08 15:06:31
*/
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
// 分
ListNode pre = head, slow = head, fast = head;
while (fast != null && fast.next != null) {
pre = slow;
slow = slow.next;
fast = fast.next.next;
}
pre.next = null; // 将链表切成两段
// 治
ListNode r1 = sortList(head);
ListNode r2 = sortList(slow);
// 合
return merge(r1, r2);
}
private ListNode merge(ListNode l1, ListNode l2) {
ListNode result = new ListNode(0);
ListNode cur = result;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
cur.next = l1;
l1 = l1.next;
} else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
if (l1 != null) {
cur.next = l1;
}
if (l2 != null) {
cur.next = l2;
}
return result.next;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-11-03 20:28:50
*/
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode slow = head, fast = head;
// 注意：这里已经在前面判断过 head == null，所以，肯定 fast != null
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode right = sortList(slow.next);
slow.next = null;
ListNode left = sortList(head);
ListNode dummy = new ListNode();
ListNode curr = dummy;
while (left != null && right != null) {
if (left.val < right.val) {
curr.next = left;
left = left.next;
} else {
curr.next = right;
right = right.next;
}
curr = curr.next;
}
curr.next = left != null ? left : right;
return dummy.next;
}