feat: add python and java solutions to leetcode problem: No.0021

Author: yanglbme

README.md

@@ -58,6 +58,7 @@
 1. [从尾到头打印链表](/lcof/面试题06.%20从尾到头打印链表/README.md)
 1. [删除链表的节点](/lcof/面试题18.%20删除链表的节点/README.md)
 1. [链表中倒数第 k 个节点](/lcci/02.02.Kth%20Node%20From%20End%20of%20List/README.md)
+1. [合并两个有序链表](/solution/0000-0099/0021.Merge%20Two%20Sorted%20Lists/README.md)
 1. [反转链表](/solution/0200-0299/0206.Reverse%20Linked%20List/README.md)
 1. [回文链表](/solution/0200-0299/0234.Palindrome%20Linked%20List/README.md)
 1. [环形链表](/solution/0100-0199/0141.Linked%20List%20Cycle/README.md)

README_EN.md

@@ -56,6 +56,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 
 1. [Delete Node in a Linked List](/solution/0200-0299/0237.Delete%20Node%20in%20a%20Linked%20List/README_EN.md)
 1. [Kth Node From End of List](/lcci/02.02.Kth%20Node%20From%20End%20of%20List/README_EN.md)
+1. [Merge Two Sorted Lists](/solution/0000-0099/0021.Merge%20Two%20Sorted%20Lists/README_EN.md)
 1. [Reverse Linked List](/solution/0200-0299/0206.Reverse%20Linked%20List/README_EN.md)
 1. [Palindrome Linked List](/solution/0200-0299/0234.Palindrome%20Linked%20List/README_EN.md)
 1. [Linked List Cycle](/solution/0100-0199/0141.Linked%20List%20Cycle/README_EN.md)

lcof/面试题25. 合并两个排序的链表/README.md

@@ -32,24 +32,18 @@
 
 class Solution:
     def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
-        if l1 is None or l2 is None:
-            return l1 or l2
-        head = ListNode(0)
-        p = head
+        dummy = ListNode()
+        cur = dummy
         while l1 and l2:
-            if l1.val < l2.val:
-                t = l1.next
-                p.next = l1
-                p = l1
-                l1 = t
+            if l1.val <= l2.val:
+                cur.next = l1
+                l1 = l1.next
             else:
-                t = l2.next
-                p.next = l2
-                p = l2
-                l2 = t
-
-        p.next = l1 or l2
-        return head.next
+                cur.next = l2
+                l2 = l2.next
+            cur = cur.next
+        cur.next = l1 or l2
+        return dummy.next
 ```
 
 ### **Java**
@@ -65,30 +59,20 @@ class Solution:
  */
 class Solution {
     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
-        if (l1 == null) {
-            return l2;
-        }
-        if (l2 == null) {
-            return l1;
-        }
-
-        ListNode head = new ListNode(0);
-        ListNode p = head;
+        ListNode dummy = new ListNode(0);
+        ListNode cur = dummy;
         while (l1 != null && l2 != null) {
-            if (l1.val < l2.val) {
-                ListNode t = l1.next;
-                p.next = l1;
-                p = l1;
-                l1 = t;
+            if (l1.val <= l2.val) {
+                cur.next = l1;
+                l1 = l1.next;
             } else {
-                ListNode t = l2.next;
-                p.next = l2;
-                p = l2;
-                l2 = t;
+                cur.next = l2;
+                l2 = l2.next;
             }
+            cur = cur.next;
         }
-        p.next = l1 == null ? l2 : l1;
-        return head.next;
+        cur.next = l1 == null ? l2 : l1;
+        return dummy.next;
     }
 }
 ```

lcof/面试题25. 合并两个排序的链表/Solution.java

@@ -8,29 +8,19 @@
  */
 class Solution {
     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
-        if (l1 == null) {
-            return l2;
-        }
-        if (l2 == null) {
-            return l1;
-        }
-
-        ListNode head = new ListNode(0);
-        ListNode p = head;
+        ListNode dummy = new ListNode(0);
+        ListNode cur = dummy;
         while (l1 != null && l2 != null) {
-            if (l1.val < l2.val) {
-                ListNode t = l1.next;
-                p.next = l1;
-                p = l1;
-                l1 = t;
+            if (l1.val <= l2.val) {
+                cur.next = l1;
+                l1 = l1.next;
             } else {
-                ListNode t = l2.next;
-                p.next = l2;
-                p = l2;
-                l2 = t;
+                cur.next = l2;
+                l2 = l2.next;
             }
+            cur = cur.next;
         }
-        p.next = l1 == null ? l2 : l1;
-        return head.next;
+        cur.next = l1 == null ? l2 : l1;
+        return dummy.next;
     }
 }

lcof/面试题25. 合并两个排序的链表/Solution.py

@@ -6,21 +6,15 @@
 
 class Solution:
     def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
-        if l1 is None or l2 is None:
-            return l1 or l2
-        head = ListNode(0)
-        p = head
+        dummy = ListNode()
+        cur = dummy
         while l1 and l2:
-            if l1.val < l2.val:
-                t = l1.next
-                p.next = l1
-                p = l1
-                l1 = t
+            if l1.val <= l2.val:
+                cur.next = l1
+                l1 = l1.next
             else:
-                t = l2.next
-                p.next = l2
-                p = l2
-                l2 = t
-
-        p.next = l1 or l2
-        return head.next
+                cur.next = l2
+                l2 = l2.next
+            cur = cur.next
+        cur.next = l1 or l2
+        return dummy.next

solution/0000-0099/0021.Merge Two Sorted Lists/README.md

@@ -24,15 +24,60 @@
 <!-- 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, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        dummy = ListNode()
+        cur = dummy
+        while l1 and l2:
+            if l1.val <= l2.val:
+                cur.next = l1
+                l1 = l1.next
+            else:
+                cur.next = l2
+                l2 = l2.next
+            cur = cur.next
+        cur.next = l1 or l2
+        return dummy.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() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        ListNode dummy = new ListNode(0);
+        ListNode cur = dummy;
+        while (l1 != null && l2 != null) {
+            if (l1.val <= l2.val) {
+                cur.next = l1;
+                l1 = l1.next;
+            } else {
+                cur.next = l2;
+                l2 = l2.next;
+            }
+            cur = cur.next;
+        }
+        cur.next = l1 == null ? l2 : l1;
+        return dummy.next;
+    }
+}
 ```
 
 ### **...**

solution/0000-0099/0021.Merge Two Sorted Lists/README_EN.md

@@ -25,13 +25,58 @@
 ### **Python3**
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        dummy = ListNode()
+        cur = dummy
+        while l1 and l2:
+            if l1.val <= l2.val:
+                cur.next = l1
+                l1 = l1.next
+            else:
+                cur.next = l2
+                l2 = l2.next
+            cur = cur.next
+        cur.next = l1 or l2
+        return dummy.next
 ```
 
 ### **Java**
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        ListNode dummy = new ListNode(0);
+        ListNode cur = dummy;
+        while (l1 != null && l2 != null) {
+            if (l1.val <= l2.val) {
+                cur.next = l1;
+                l1 = l1.next;
+            } else {
+                cur.next = l2;
+                l2 = l2.next;
+            }
+            cur = cur.next;
+        }
+        cur.next = l1 == null ? l2 : l1;
+        return dummy.next;
+    }
+}
 ```
 
 ### **...**

solution/0000-0099/0021.Merge Two Sorted Lists/Solution.java

@@ -3,25 +3,26 @@
  * public class ListNode {
  *     int val;
  *     ListNode next;
- *     ListNode(int x) { val = x; }
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
  * }
  */
 class Solution {
     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
-        ListNode head = new ListNode(0);
-        head.next = null;
-        ListNode current = head;
+        ListNode dummy = new ListNode(0);
+        ListNode cur = dummy;
         while (l1 != null && l2 != null) {
             if (l1.val <= l2.val) {
-                current.next = l1;
+                cur.next = l1;
                 l1 = l1.next;
             } else {
-                current.next = l2;
+                cur.next = l2;
                 l2 = l2.next;
             }
-            current = current.next;
+            cur = cur.next;
         }
-        current.next = l1 != null ? l1 : l2;
-        return head.next;
+        cur.next = l1 == null ? l2 : l1;
+        return dummy.next;
     }
 }

solution/0000-0099/0021.Merge Two Sorted Lists/Solution.py

@@ -1,21 +1,19 @@
 # Definition for singly-linked list.
 # class ListNode:
-#     def __init__(self, x):
-#         self.val = x
-#         self.next = None
-
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
 class Solution:
-    def mergeTwoLists(self, l1, l2):
-        """
-        :type l1: ListNode
-        :type l2: ListNode
-        :rtype: ListNode
-        """
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        dummy = ListNode()
+        cur = dummy
         while l1 and l2:
-            if l1.val < l2.val:
-                l1.next=self.mergeTwoLists(l1.next,l2)
-                return l1
+            if l1.val <= l2.val:
+                cur.next = l1
+                l1 = l1.next
             else:
-                l2.next=self.mergeTwoLists(l1,l2.next)
-                return l2
-        return l1 or l2
+                cur.next = l2
+                l2 = l2.next
+            cur = cur.next
+        cur.next = l1 or l2
+        return dummy.next