feat(ml): $2.add-two-numbers.md

problems/2.add-two-numbers.md

@@ -48,11 +48,11 @@ https://leetcode-cn.com/problems/add-two-numbers/
 
 ## 代码
 
-- 语言支持：JS，C++
+- 语言支持：JS，C++，Java，Python
 
-JavaScript:
+JavaScript Code:
 
-```js
+```js 
 /**
  * Definition for singly-linked list.
  * function ListNode(val) {
@@ -102,7 +102,7 @@ var addTwoNumbers = function (l1, l2) {
 };
 ```
 
-C++
+C++ Code:
 
 > C++代码与上面的 JavaScript 代码略有不同：将 carry 是否为 0 的判断放到了 while 循环中
 
@@ -141,6 +141,82 @@ public:
 };
 ```
 
+
+Java Code:
+
+```java
+class Solution {
+    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
+        ListNode dummyHead = new ListNode(0);
+        ListNode cur = dummyHead;
+        int carry = 0;
+
+        while(l1 != null || l2 != null)
+        {
+            int sum = carry;
+            if(l1 != null)
+            {
+                sum += l1.val;
+                l1 = l1.next;
+            }
+            if(l2 != null)
+            {
+                sum += l2.val;
+                l2 = l2.next;
+            }
+            // 创建新节点
+            carry = sum / 10;
+            cur.next = new ListNode(sum % 10);
+            cur = cur.next;
+
+        }
+        if (carry > 0) {
+            cur.next = new ListNode(carry);
+        }
+        return dummyHead.next;
+    }
+}
+
+```
+
+
+Python Code:
+
+```py
+class Solution:
+    def addTwoNumbers(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """
+        res=ListNode(0)
+        head=res
+        carry=0
+        while l1 or l2 or carry!=0:
+            sum=carry
+            if l1:
+                sum+=l1.val
+                l1=l1.next
+            if l2:
+                sum+=l2.val
+                l2=l2.next
+            # set value
+            if sum<=9:
+                res.val=sum
+                carry=0
+            else:
+                res.val=sum%10
+                carry=sum//10
+            # creat new node
+            if l1 or l2 or carry!=0:
+                res.next=ListNode(0)
+                res=res.next
+        return head
+
+```
+
+
 **复杂度分析**
 
 - 时间复杂度：$O(N)$