feixiangcode
diff --git a/‎Week_01/id_126/LeetCode_21_126.java‎
Lines changed: 41 additions & 0 deletions b/‎Week_01/id_126/LeetCode_21_126.java‎
Lines changed: 41 additions & 0 deletions
diff --git a/‎Week_01/id_126/NOTE.md‎
Lines changed: 5 additions & 1 deletion b/‎Week_01/id_126/NOTE.md‎
Lines changed: 5 additions & 1 deletion
diff --git a/‎Week_01/id_126/test.txt‎ b/‎Week_01/id_126/test.txt‎
@@ -0,0 +1,41 @@
+/**
+ * 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) {
+        //创建一个新的listnode
+        //两个有序的listnode每个取一个，比较，并加入新的Listnode
+        ListNode listNode = new ListNode(0);
+        ListNode current = listNode;
+        while (l1 != null && l2 != null) {
+            if (l1.val < l2.val) {
+                //l1加入listnode  l1位置后移
+                current.next = l1;
+                current = l1;
+                l1 = l1.next;
+
+            } else {
+                //l2加入listnode  l2位置后移
+                current.next = l2;
+                current = l2;
+                l2 = l2.next;
+
+            }
+
+        }
+
+        if (l1 == null) {
+            current.next = l2;
+        }
+        if (l2 == null) {
+            current.next = l1;
+        }
+
+        return listNode.next;
+    }
+}
@@ -1 +1,5 @@
-# 学习笔记
+# 学习笔记
+
+解题思路：
+技巧一：初始化的时候把head节点初始化一个给定的值，这样，后续操作就不用考虑头节点问题，直接排序好了，把最小的赋值给head.next就可以了。
+技巧二：两个链表为空和不为空的判断，循环逻辑依赖判断条件