diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md
index 643f6055..59ded523 100644
--- a/problems/0024.两两交换链表中的节点.md
+++ b/problems/0024.两两交换链表中的节点.md
@@ -129,6 +129,23 @@ class Solution {
 ```
 
 Python：
+```python
+class Solution:
+    def swapPairs(self, head: ListNode) -> ListNode:
+        dummy = ListNode(0) #设置一个虚拟头结点
+        dummy.next = head
+        cur = dummy
+        while cur.next and cur.next.next: 
+            tmp = cur.next #记录临时节点
+            tmp1 = cur.next.next.next #记录临时节点
+            
+            cur.next = cur.next.next          #步骤一
+            cur.next.next = tmp               #步骤二
+            cur.next.next.next = tmp1         #步骤三
+            
+            cur = cur.next.next #cur移动两位，准备下一轮交换
+        return dummy.next
+```
 
 Go：
 ```go
diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md
index 396231fe..d1bccd64 100644
--- a/problems/0202.快乐数.md
+++ b/problems/0202.快乐数.md
@@ -108,7 +108,29 @@ class Solution {
 ```
 
 Python：
-
+```python
+class Solution:
+    def isHappy(self, n: int) -> bool:
+        set_ = set()
+        while 1:
+            sum_ = self.getSum(n)
+            if sum_ == 1:
+                return True
+            #如果这个sum曾经出现过，说明已经陷入了无限循环了，立刻return false
+            if sum_ in set_:
+                return False
+            else:
+                set_.add(sum_)
+            n = sum_
+            
+    #取数值各个位上的单数之和
+    def getSum(self, n):
+        sum_ = 0
+        while n > 0:
+            sum_ += (n%10) * (n%10)
+            n //= 10
+        return sum_
+```
 
 Go：
 ```go
diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md
index 6a6faf63..ba942f70 100644
--- a/problems/0242.有效的字母异位词.md
+++ b/problems/0242.有效的字母异位词.md
@@ -113,7 +113,22 @@ class Solution {
 ```
 
 Python：
-
+```python
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        record = [0] * 26
+        for i in range(len(s)):
+            #并不需要记住字符a的ASCII，只要求出一个相对数值就可以了
+            record[ord(s[i]) - ord("a")] += 1
+        print(record)
+        for i in range(len(t)):
+            record[ord(t[i]) - ord("a")] -= 1
+        for i in range(26):
+            if record[i] != 0:
+                #record数组如果有的元素不为零0，说明字符串s和t 一定是谁多了字符或者谁少了字符。
+                return False
+        return True
+```
 
 Go：
 ```go