add: 合并两个有序链表

README.md

@@ -271,3 +271,8 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
   - LeetCode 23. 合并K个排序链表 https://leetcode-cn.com/problems/merge-k-sorted-lists/
   - LintCode 104. 合并k个排序链表 https://www.lintcode.com/problem/merge-k-sorted-lists/description
+
+- [合并两个有序链表](src/list/merge-two-sorted-lists.js)
+
+  - 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

src/list/merge-k-sorted-lists.js

@@ -4,8 +4,8 @@
  * @param {ListNode[]} right
  */
 export const merge = (left, right) => {
-  const head = {}
+  const dummyHead = {}
-  let current = head
+  let current = dummyHead
   while (left !== null && right !== null) {
     if (left.val < right.val) {
       if (left.val !== null) {
@@ -28,7 +28,7 @@ export const merge = (left, right) => {
     current.next = left
   }
 
-  return head.next
+  return dummyHead.next
 }
 
 /**

src/list/merge-two-sorted-lists.js

@@ -0,0 +1,42 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} l1
+ * @param {ListNode} l2
+ * @return {ListNode}
+ */
+export const mergeTwoLists = function (l1, l2) {
+  // src/list/merge-k-sorted-lists.js 中已经写过了
+  const dummyHead = {}
+  let current = dummyHead
+  while (l1 !== null && l2 !== null) {
+    if (l1.val < l2.val) {
+      if (l1.val !== null) {
+        current.next = l1
+        current = current.next
+      }
+
+      l1 = l1.next
+    } else {
+      if (l2.val !== null) {
+        current.next = l2
+        current = current.next
+      }
+
+      l2 = l2.next
+    }
+  }
+
+  if (l1 === null) {
+    current.next = l2
+  } else {
+    current.next = l1
+  }
+
+  return dummyHead.next
+}

test/list/merge-two-sorted-lists.test.js

@@ -0,0 +1,23 @@
+import { mergeTwoLists } from '../../src/list/merge-two-sorted-lists'
+
+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('合并两个有序链表', () => {
+  expect(mergeTwoLists(arrToList([1, 2, 4]), arrToList([1, 3, 4]))).toEqual(arrToList([1, 1, 2, 3, 4, 4]))
+  expect(mergeTwoLists(arrToList([null]), arrToList([0, 3, 3, null]))).toEqual(arrToList([0, 3, 3, null]))
+  expect(mergeTwoLists(arrToList([1, 3, 8, 11, 15, null]), arrToList([2, null]))).toEqual(arrToList([1, 2, 3, 8, 11, 15, null]))
+})