add: 链表排序

README.md

@@ -276,3 +276,8 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
   - LeetCode 21. 合并两个有序链表 https://leetcode-cn.com/problems/merge-two-sorted-lists/
   - LintCode 165. 合并两个排序链表 https://www.lintcode.com/problem/merge-two-sorted-lists/description
+
+- [链表排序](src/list/sort-list.js)
+
+  - LeetCode 148. 排序链表 https://leetcode-cn.com/problems/sort-list/
+  - LintCode 98. 链表排序 https://www.lintcode.com/problem/sort-list/description

src/list/sort-list.js

@@ -0,0 +1,31 @@
+import { mergeTwoLists } from './merge-two-sorted-lists'
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+export const sortList = function (head) {
+  if (head === null || head.next === null) return head
+
+  let slow = head
+  let fast = head
+  while (fast !== null) {
+    fast = fast.next
+    fast = fast !== null ? fast.next : null
+    if (fast !== null) {
+      slow = slow.next
+    }
+  }
+  const half = slow.next
+  slow.next = null
+  const left = sortList(head)
+  const right = sortList(half)
+  return mergeTwoLists(left, right)
+}

test/list/sort-list.test.js

@@ -0,0 +1,31 @@
+import { sortList } from '../../src/list/sort-list'
+
+function ListNode (val) {
+  this.val = val
+  this.next = null
+}
+
+const arrToList = (arr) => {
+  const head = new ListNode(arr[0])
+  let current = head
+  for (let n = 1, len = arr.length; n < len; n++) {
+    current.next = new ListNode(arr[n])
+    current = current.next
+  }
+
+  return head
+}
+
+test('链表排序', () => {
+  const source = [4, 2, 1, 3]
+  expect(sortList(arrToList(source))).toEqual(arrToList(source.sort((a, b) => a - b)))
+
+  const source2 = [-1, 5, 3, 4, 0]
+  expect(sortList(arrToList(source2))).toEqual(arrToList(source2.sort((a, b) => a - b)))
+
+  const source3 = [1, 3, 2, null]
+  expect(sortList(arrToList(source3))).toEqual(arrToList([1, 2, 3]))
+
+  const source4 = [1, 7, 2, 6, null]
+  expect(sortList(arrToList(source4))).toEqual(arrToList([1, 2, 6, 7]))
+})