feat: add python and java solutions to lcci question

添加《程序员面试金典题解》：面试题 02.05. 链表求和

Author: yanglbme

lcci/02.05.Sum Lists/README.md

@@ -27,20 +27,64 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+同时遍历两链表，求节点的和与进位。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
+        p = ListNode(-1)
+        carry, t = 0, p
+        while l1 or l2:
+            s = (0 if l1 is None else l1.val) + (0 if l2 is None else l2.val) + carry
+            carry = 1 if s > 9 else 0
+            t.next = ListNode(s % 10)
+            t = t.next
+            l1 = l1.next if l1 else l1
+            l2 = l2.next if l2 else l2
+        t.next = None if carry == 0 else ListNode(carry)
+        return p.next
+
 
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
+        ListNode p = new ListNode(-1);
+        int carry = 0;
+        ListNode t = p;
+        while (l1 != null || l2 != null) {
+            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
+            t.next = new ListNode(s % 10);
+            carry = s > 9 ? 1 : 0;
+            t = t.next;
+            l1 = l1 == null ? l1 : l1.next;
+            l2 = l2 == null ? l2 : l2.next;
+        }
+        t.next = carry == 0 ? null : new ListNode(carry);
+        return p.next;
+    }
+}
 ```
 
 ### ...

lcci/02.05.Sum Lists/README_EN.md

@@ -1,44 +1,107 @@
-# [02.05. Sum Lists](https://leetcode-cn.com/problems/sum-lists-lcci)
-
-## Description
-<p>You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1&#39;s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.</p>
-
-<p>&nbsp;</p>
-
-<p><strong>Example: </strong></p>
-
-<pre>
-<strong>Input: </strong>(7 -&gt; 1 -&gt; 6) + (5 -&gt; 9 -&gt; 2). That is, 617 + 295.
-<strong>Output: </strong>2 -&gt; 1 -&gt; 9. That is, 912.
-</pre>
-
-<p><strong>Follow Up:&nbsp;</strong>Suppose the digits are stored in forward order. Repeat the above problem.</p>
-
-<p><strong>Example: </strong></p>
-
-<pre>
-<strong>Input: </strong>(6 -&gt; 1 -&gt; 7) + (2 -&gt; 9 -&gt; 5). That is, 617 + 295.
-<strong>Output: </strong>9 -&gt; 1 -&gt; 2. That is, 912.
-</pre>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [02.05. Sum Lists](https://leetcode-cn.com/problems/sum-lists-lcci)
+
+## Description
+<p>You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1&#39;s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.</p>
+
+
+
+<p>&nbsp;</p>
+
+
+
+<p><strong>Example: </strong></p>
+
+
+
+<pre>
+
+<strong>Input: </strong>(7 -&gt; 1 -&gt; 6) + (5 -&gt; 9 -&gt; 2). That is, 617 + 295.
+
+<strong>Output: </strong>2 -&gt; 1 -&gt; 9. That is, 912.
+
+</pre>
+
+
+
+<p><strong>Follow Up:&nbsp;</strong>Suppose the digits are stored in forward order. Repeat the above problem.</p>
+
+
+
+<p><strong>Example: </strong></p>
+
+
+
+<pre>
+
+<strong>Input: </strong>(6 -&gt; 1 -&gt; 7) + (2 -&gt; 9 -&gt; 5). That is, 617 + 295.
+
+<strong>Output: </strong>9 -&gt; 1 -&gt; 2. That is, 912.
+
+</pre>
+
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
+        p = ListNode(-1)
+        carry, t = 0, p
+        while l1 or l2:
+            s = (0 if l1 is None else l1.val) + (0 if l2 is None else l2.val) + carry
+            carry = 1 if s > 9 else 0
+            t.next = ListNode(s % 10)
+            t = t.next
+            l1 = l1.next if l1 else l1
+            l2 = l2.next if l2 else l2
+        t.next = None if carry == 0 else ListNode(carry)
+        return p.next
+
+
+```
+
+### Java
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
+        ListNode p = new ListNode(-1);
+        int carry = 0;
+        ListNode t = p;
+        while (l1 != null || l2 != null) {
+            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
+            t.next = new ListNode(s % 10);
+            carry = s > 9 ? 1 : 0;
+            t = t.next;
+            l1 = l1 == null ? l1 : l1.next;
+            l2 = l2 == null ? l2 : l2.next;
+        }
+        t.next = carry == 0 ? null : new ListNode(carry);
+        return p.next;
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/02.05.Sum Lists/Solution.java

@@ -0,0 +1,25 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
+        ListNode p = new ListNode(-1);
+        int carry = 0;
+        ListNode t = p;
+        while (l1 != null || l2 != null) {
+            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
+            t.next = new ListNode(s % 10);
+            carry = s > 9 ? 1 : 0;
+            t = t.next;
+            l1 = l1 == null ? l1 : l1.next;
+            l2 = l2 == null ? l2 : l2.next;
+        }
+        t.next = carry == 0 ? null : new ListNode(carry);
+        return p.next;
+    }
+}

lcci/02.05.Sum Lists/Solution.py

@@ -0,0 +1,21 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+
+class Solution:
+    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
+        p = ListNode(-1)
+        carry, t = 0, p
+        while l1 or l2:
+            s = (0 if l1 is None else l1.val) + \
+                (0 if l2 is None else l2.val) + carry
+            carry = 1 if s > 9 else 0
+            t.next = ListNode(s % 10)
+            t = t.next
+            l1 = l1.next if l1 else l1
+            l2 = l2.next if l2 else l2
+        t.next = None if carry == 0 else ListNode(carry)
+        return p.next