Merge two sorted-lists add solution

Code by Java,C++

Author: doublehearts

problems/21.merge-two-sorted-lists.md

@@ -44,7 +44,7 @@ https://leetcode-cn.com/problems/merge-two-sorted-lists
 
 ## 代码
 
-- 语言支持：CPP, JS
+- 语言支持：CPP, JS, Java, Python
 
 CPP Code:
 
@@ -99,6 +99,45 @@ const mergeTwoLists = function (l1, l2) {
 };
 ```
 
+Java Code:
+
+```java
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        if (l1 == null) {
+            return l2;
+        }
+        else if (l2 == null) {
+            return l1;
+        }
+        else if (l1.val < l2.val) {
+            l1.next = mergeTwoLists(l1.next, l2);
+            return l1;
+        }
+        else {
+            l2.next = mergeTwoLists(l1, l2.next);
+            return l2;
+        }
+
+    }
+}
+```
+
+Python Code:
+
+```py
+class Solution:
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        if not l1: return l2  # 终止条件，直到两个链表都空
+        if not l2: return l1
+        if l1.val <= l2.val:  # 递归调用
+            l1.next = self.mergeTwoLists(l1.next,l2)
+            return l1
+        else:
+            l2.next = self.mergeTwoLists(l1,l2.next)
+            return l2
+```
+
 **复杂度分析**
 
 M、N 是两条链表 l1、l2 的长度
@@ -156,6 +195,56 @@ var mergeTwoLists = function (l1, l2) {
 };
 ```
 
+迭代的Java代码如下：
+
+```java
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        ListNode prehead = new ListNode(-1);
+
+        ListNode prev = prehead;
+        while (l1 != null && l2 != null) {
+            if (l1.val <= l2.val) {
+                prev.next = l1;
+                l1 = l1.next;
+            } else {
+                prev.next = l2;
+                l2 = l2.next;
+            }
+            prev = prev.next;
+        }
+
+        // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
+        prev.next = l1 == null ? l2 : l1;
+
+        return prehead.next;
+    }
+}
+```
+
+迭代的Python代码如下：
+
+```py
+class Solution:
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        prehead = ListNode(-1)
+
+        prev = prehead
+        while l1 and l2:
+            if l1.val <= l2.val:
+                prev.next = l1
+                l1 = l1.next
+            else:
+                prev.next = l2
+                l2 = l2.next            
+            prev = prev.next
+
+        # 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
+        prev.next = l1 if l1 is not None else l2
+
+        return prehead.next
+```
+
 大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：https://github.com/azl397985856/leetcode 。 目前已经 40K star 啦。
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