diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md
index 6a178a25..145f64aa 100644
--- a/problems/0474.一和零.md
+++ b/problems/0474.一和零.md
@@ -210,19 +210,35 @@ class Solution {
 ```
 
 ### Python
+DP（版本一）
 ```python
 class Solution:
     def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
-        dp = [[0] * (n + 1) for _ in range(m + 1)]	# 默认初始化0
+        dp = [[0] * (n + 1) for _ in range(m + 1)]  # 创建二维动态规划数组，初始化为0
+        for s in strs:  # 遍历物品
+            zeroNum = s.count('0')  # 统计0的个数
+            oneNum = len(s) - zeroNum  # 统计1的个数
+            for i in range(m, zeroNum - 1, -1):  # 遍历背包容量且从后向前遍历
+                for j in range(n, oneNum - 1, -1):
+                    dp[i][j] = max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1)  # 状态转移方程
+        return dp[m][n]
+
+```
+DP（版本二）
+```python
+class Solution:
+    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
+        dp = [[0] * (n + 1) for _ in range(m + 1)]  # 创建二维动态规划数组，初始化为0
         # 遍历物品
-        for str in strs:
-            ones = str.count('1')
-            zeros = str.count('0')
-            # 遍历背包容量且从后向前遍历！
+        for s in strs:
+            ones = s.count('1')  # 统计字符串中1的个数
+            zeros = s.count('0')  # 统计字符串中0的个数
+            # 遍历背包容量且从后向前遍历
             for i in range(m, zeros - 1, -1):
                 for j in range(n, ones - 1, -1):
-                    dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)
+                    dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)  # 状态转移方程
         return dp[m][n]
+
 ```
 
 ### Go