21题

Author: changhongyuan

.idea/leetcode/editor.xml

@@ -52,12 +52,28 @@ class MergeTwoSortedLists{
  */
 class Solution {
     public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
-        return null;
+        if (null == list1) {
+            return list2;
+        }
+        if (null == list2) {
+            return list1;
+        }
+        if (list1.val < list2.val) {
+            list1.next = mergeTwoLists(list1.next, list2);
+            return list1;
+        } else {
+            list2.next = mergeTwoLists(list2.next, list1);
+            return list2;
+        }
     }
 }
 //leetcode submit region end(Prohibit modification and deletion)
 
     public static void main(String[] args) {
         Solution solution = new MergeTwoSortedLists().new Solution();
+        ListNode list1 = new ListNode(1, new ListNode(2, new ListNode(4)));
+        ListNode list2 = new ListNode(1, new ListNode(3, new ListNode(4)));
+        ListNode result = solution.mergeTwoLists(list1, list2);
+        System.out.println(result.toString());
     }
 }

src/main/java/com/changhongyuan/leetcode/editor/cn/[21]合并两个有序链表.java

@@ -0,0 +1,79 @@
+### 思路
+
+- 标签：链表、递归
+- 这道题可以使用递归实现，新链表也不需要构造新节点，我们下面列举递归三个要素
+- 终止条件：两条链表分别名为 `l1` 和 `l2`，当 `l1` 为空或 `l2` 为空时结束
+- 返回值：每一层调用都返回排序好的链表头
+- 本级递归内容：如果 `l1` 的 `val` 值更小，则将 `l1.next` 与排序好的链表头相接，`l2` 同理
+- $O(m+n)$，$m$ 为 `l1`的长度，$n$ 为 `l2` 的长度
+
+### 代码
+
+* []
+
+```Java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        if(l1 == null) {
+            return l2;
+        }
+        if(l2 == null) {
+            return l1;
+        }
+
+        if(l1.val < l2.val) {
+            l1.next = mergeTwoLists(l1.next, l2);
+            return l1;
+        } else {
+            l2.next = mergeTwoLists(l1, l2.next);
+            return l2;
+        }
+    }
+}
+```
+
+* []
+
+```JavaScript
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} l1
+ * @param {ListNode} l2
+ * @return {ListNode}
+ */
+var mergeTwoLists = function(l1, l2) {
+    if(l1 === null){
+        return l2;
+    }
+    if(l2 === null){
+        return l1;
+    }
+    if(l1.val < l2.val){
+        l1.next = mergeTwoLists(l1.next, l2);
+        return l1;
+    }else{
+        l2.next = mergeTwoLists(l1, l2.next);
+        return l2;
+    }
+};
+```
+
+### 画解
+
+<![frame_00001.png](https://pic.leetcode-cn.com/7ddaf1beb64fdef4393cc6ebd0dfd1723b97d2c183ab5c8414c0898027623a00-frame_00001.png),![frame_00002.png](https://pic.leetcode-cn.com/f4b7e354473d2bf28283ac3c410bc81e9f7ecb35f14189de9fadc041452c2653-frame_00002.png),![frame_00003.png](https://pic.leetcode-cn.com/001e4c2fdd8b5d725bc25df6373f7590404d9ef16efdea6e3700b68c23500a7a-frame_00003.png),![frame_00004.png](https://pic.leetcode-cn.com/5fbc72d56f32a8b1bc34db4bbd1588abebb4942348d8ea22fdb60724c8e4986c-frame_00004.png),![frame_00007.png](https://pic.leetcode-cn.com/912a9fef02ca9d5b4cdb891c2500f496a5f329adafa22f8ecdfd1cb591434b92-frame_00007.png)>
+
+想看大鹏画解更多高频面试题，欢迎阅读大鹏的 LeetBook：[《画解剑指 Offer 》](https://leetcode-cn.com/leetbook/detail/illustrate-lcof/)，O(∩_∩)O