solve problem Merge Two Sorted Lists

Author: P-ppc

README.md

@@ -68,6 +68,7 @@ All solutions will be accepted!
 |806|[Number Of Lines To Write String](https://leetcode-cn.com/problems/number-of-lines-to-write-string/description/)|[java/py/js](./algorithms/NumberOfLinesToWriteString)|Easy|
 |566|[Reshape The Matrix](https://leetcode-cn.com/problems/reshape-the-matrix/description/)|[java/py/js](./algorithms/ReshapeTheMatrix)|Easy|
 |575|[Distribute Candies](https://leetcode-cn.com/problems/distribute-candies/description/)|[java/py/js](./algorithms/DistributeCandies)|Easy|
+|21|[Merge Two Sorted Lists](https://leetcode-cn.com/problems/merge-two-sorted-lists/description/)|[java/py/js](./algorithms/MergeTwoSortedLists)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/MergeTwoSortedLists/README.md

@@ -0,0 +1,2 @@
+# Merge Two Sorted Lists
+This problem is easy to solve by two linked list

algorithms/MergeTwoSortedLists/Solution.java

@@ -0,0 +1,31 @@
+/**
+ * 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) {
+        ListNode res = null,
+                 head = null;
+        while (l1 != null && l2 != null) {
+            if (res == null) {
+                head = res = new ListNode(Math.min(l1.val, l2.val));
+            } else {
+                head.next = new ListNode(Math.min(l1.val, l2.val));
+                head = head.next;
+            }
+            if (l1.val <= l2.val) {
+                l1 = l1.next;
+            } else {
+                l2 = l2.next;
+            }
+        }
+        
+        if (res == null) res = l1 != null ? l1 : l2;
+        else head.next = l1 != null ? l1 : l2;
+        return res;
+    }
+}

algorithms/MergeTwoSortedLists/solution.js

@@ -0,0 +1,37 @@
+/**
+ * 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) {
+    let head = res = null
+    
+    while (l1 !== null && l2 !== null) {
+        if (res === null) {
+            head = res = new ListNode(Math.min(l1.val, l2.val))
+        } else {
+            head.next = new ListNode(Math.min(l1.val, l2.val))
+            head = head.next
+        }
+        if (l1.val <= l2.val) {
+            l1 = l1.next
+        } else {
+            l2 = l2.next
+        }
+    }
+    
+    if (res === null) {
+        res = l1 !== null ? l1 : l2
+    } else {
+        head.next = l1 !== null ? l1 : l2
+    }
+    
+    return res
+};

algorithms/MergeTwoSortedLists/solution.py

@@ -0,0 +1,35 @@
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def mergeTwoLists(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """        
+        res = None
+        head = None
+        while l1 and l2:
+            if res == None:
+                res = ListNode(min(l1.val, l2.val))
+                head = res
+            else:
+                head.next = ListNode(min(l1.val, l2.val))
+                head = head.next
+            if l1.val <= l2.val:
+                l1 = l1.next
+            else:
+                l2 = l2.next
+        if l1 and res:
+            head.next = l1
+        elif l2 and res:
+            head.next = l2
+        elif l1 and not res:
+            res = l1
+        elif l2 and not res:
+            res = l2
+        return res