104 lines
2.3 KiB
Markdown
104 lines
2.3 KiB
Markdown
# 排序链表
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## 题目
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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 。
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示例 1：
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输入：head = [4,2,1,3]
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输出：[1,2,3,4]
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示例 2：
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输入：head = [-1,5,3,4,0]
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输出：[-1,0,3,4,5]
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示例 3：
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输入：head = []
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输出：[]
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提示：
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链表中节点的数目在范围 [0, 5 * 104] 内
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-105 <= Node.val <= 105
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进阶：你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？
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## 思路
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- 归并排序
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- 拆分
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- 快慢指针
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- 合并
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## 代码
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```java
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode() {}
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* ListNode(int val) { this.val = val; }
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* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
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* }
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*/
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class Solution {
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public ListNode sortList(ListNode head) {
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// -- Divide -> Sort -> Merge --
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// Special
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if (head == null || head.next == null) {
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return head;
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}
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// Init: slow, fast, dummyHead
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ListNode dummyHead = new ListNode(-1);
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dummyHead.next = head;
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ListNode slow = dummyHead;
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ListNode fast = dummyHead;
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while (fast.next != null && fast.next.next != null) {
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fast = fast.next.next;
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slow = slow.next;
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}
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// Divide: [headA .... tailA(slow)] [headB(slow.next) ... tailB]
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ListNode headA = head;
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ListNode headB = slow.next;
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slow.next = null;
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// Sort
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headA = sortList(headA);
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headB = sortList(headB);
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// Merge
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return mergeList(headA, headB);
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}
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public ListNode mergeList(ListNode headA, ListNode headB) {
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// Init: dummyHead, tail
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ListNode dummyHead = new ListNode(-1);
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ListNode tail = dummyHead;
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// Traverse: Merge
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while (headA != null && headB != null) {
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int valA = headA.val;
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int valB = headB.val;
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if (valA < valB) {
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tail.next = headA;
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headA = headA.next;
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} else {
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tail.next = headB;
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headB = headB.next;
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}
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tail = tail.next;
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}
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// Operation: After
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tail.next = headA == null ? headB : headA;
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return dummyHead.next;
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}
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}
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``` |