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| 1 | +<h1 style="text-align: center;"> <span style="color: #FFB822;">148. 排序链表</span> </h1> |
| 2 | + |
| 3 | +### 🚀 LeetCode |
| 4 | + |
| 5 | +<base target="_blank"> |
| 6 | + |
| 7 | +<span style="color: #FFB822;">**Medium**</span> [**https://leetcode.cn/problems/sort-list/**](https://leetcode.cn/problems/sort-list/) |
| 8 | + |
| 9 | +--- |
| 10 | + |
| 11 | +### ❓ Description |
| 12 | + |
| 13 | +<br/> |
| 14 | + |
| 15 | +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**。 |
| 16 | + |
| 17 | +<br/> |
| 18 | + |
| 19 | +**示例 1：** |
| 20 | + |
| 21 | +<img src="../../public/0148/sort-list-1.jpg" alt="sort-list-1.jpg"/> |
| 22 | + |
| 23 | +``` |
| 24 | +输入: head = [4, 2, 1, 3] |
| 25 | +输出: [1, 2, 3, 4] |
| 26 | +``` |
| 27 | + |
| 28 | +**示例 2：** |
| 29 | + |
| 30 | +<img src="../../public/0148/sort-list-2.jpg" alt="sort-list-2.jpg"/> |
| 31 | + |
| 32 | +``` |
| 33 | +输入: head = [-1, 5, 3, 4, 0] |
| 34 | +输出: [-1, 0, 3, 4, 5] |
| 35 | +``` |
| 36 | + |
| 37 | +**示例 3：** |
| 38 | + |
| 39 | +``` |
| 40 | +输入: head = [] |
| 41 | +输出: [] |
| 42 | +``` |
| 43 | + |
| 44 | +<br/> |
| 45 | + |
| 46 | +**提示：** |
| 47 | + |
| 48 | +* 链表中节点的数目在范围 `[0, 5 * 10^4]` 内 |
| 49 | +* `-10^5 <= Node.val <= 10^5` |
| 50 | + |
| 51 | +<br/> |
| 52 | + |
| 53 | +**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？ |
| 54 | + |
| 55 | +--- |
| 56 | + |
| 57 | +### ❗ Solution |
| 58 | + |
| 59 | +<br/> |
| 60 | + |
| 61 | +#### Java |
| 62 | + |
| 63 | +``` |
| 64 | +/** |
| 65 | + * Definition for singly-linked list. |
| 66 | + * public class ListNode { |
| 67 | + * int val; |
| 68 | + * ListNode next; |
| 69 | + * ListNode() {} |
| 70 | + * ListNode(int val) { this.val = val; } |
| 71 | + * ListNode(int val, ListNode next) { this.val = val; this.next = next; } |
| 72 | + * } |
| 73 | + */ |
| 74 | +class Solution { |
| 75 | + public ListNode sortList(ListNode head) { |
| 76 | + return divide(head, null); |
| 77 | + } |
| 78 | +
|
| 79 | + // 对半拆分 |
| 80 | + public ListNode divide(ListNode head, ListNode tail) { |
| 81 | + // 如果头节点为空或者头节点的下一个节点为空 |
| 82 | + // 说明当前子链表为空或者只有一个节点, 结束递归, 开始返回 |
| 83 | + if (head == null || head.next == null) { |
| 84 | + return head; |
| 85 | + } |
| 86 | +
|
| 87 | + // 快慢指针遍历链表, 找到中间节点 |
| 88 | + ListNode slow = head, fast = head; |
| 89 | +
|
| 90 | + // 快指针走两步, 慢指针走一步 |
| 91 | + // 直到快指针到达链表尾部, 此时慢指针指向的就是中间节点 |
| 92 | + // 如果当前链表长度为偶数, 慢指针指向的是中间两个节点中, 靠左的那一个节点 |
| 93 | + while (fast != tail) { |
| 94 | + fast = fast.next; |
| 95 | + if (fast != tail) { |
| 96 | + fast = fast.next; |
| 97 | + slow = slow.next; |
| 98 | + } |
| 99 | + } |
| 100 | +
|
| 101 | + // 获取两个中间节点 |
| 102 | + ListNode mid1 = slow; |
| 103 | + ListNode mid2 = slow.next; |
| 104 | + // 切断左右子链表的连接 |
| 105 | + mid1.next = null; |
| 106 | +
|
| 107 | + // 递归, 将左右子链表继续左右拆分, 一直拆分到最小单元, 也就是直到子链表为空或者长度为 1 |
| 108 | + ListNode left = divide(head, mid1); |
| 109 | + ListNode right = divide(mid2, tail); |
| 110 | + // 合并左右子链表, 并在合并的同时进行排序 |
| 111 | + return merge(left, right); |
| 112 | + } |
| 113 | +
|
| 114 | + // 归并排序 |
| 115 | + public ListNode merge(ListNode left, ListNode right) { |
| 116 | + ListNode head = new ListNode(0); |
| 117 | + ListNode node = head; |
| 118 | + while (left != null && right != null) { |
| 119 | + if (left.val <= right.val) { |
| 120 | + node.next = left; |
| 121 | + left = left.next; |
| 122 | + } else { |
| 123 | + node.next = right; |
| 124 | + right = right.next; |
| 125 | + } |
| 126 | + node = node.next; |
| 127 | + } |
| 128 | + if (left != null) { |
| 129 | + node.next = left; |
| 130 | + } else if (right != null) { |
| 131 | + node.next = right; |
| 132 | + } |
| 133 | + return head.next; |
| 134 | + } |
| 135 | +} |
| 136 | +``` |
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