diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md
index 0664b0d9..19f4cc72 100644
--- a/problems/1143.最长公共子序列.md
+++ b/problems/1143.最长公共子序列.md
@@ -31,7 +31,7 @@
 输入：text1 = "abc", text2 = "def"      
 输出：0        
 解释：两个字符串没有公共子序列，返回 0。    
- 
+
 提示:       
 * 1 <= text1.length <= 1000
 * 1 <= text2.length <= 1000
@@ -126,12 +126,44 @@ public:
 
 ## 其他语言版本
 
-
 Java：
 
+```java
+class Solution {
+    public int longestCommonSubsequence(String text1, String text2) {
+        int[][] dp = new int[text1.length() + 1][text2.length() + 1]; // 先对dp数组做初始化操作
+        for (int i = 1 ; i <= text1.length() ; i++) {
+            char char1 = text1.charAt(i - 1);
+            for (int j = 1; j <= text2.length(); j++) {
+                char char2 = text2.charAt(j - 1);
+                if (char1 == char2) { // 开始列出状态转移方程
+                    dp[i][j] = dp[i - 1][j - 1] + 1;
+                } else {
+                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
+                }
+            }
+        }
+        return dp[text1.length()][text2.length()];
+    }
+}
+```
 
 Python：
 
+```python
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        len1, len2 = len(text1)+1, len(text2)+1
+        dp = [[0 for _ in range(len1)] for _ in range(len2)] # 先对dp数组做初始化操作
+        for i in range(1, len2):
+            for j in range(1, len1): # 开始列出状态转移方程
+                if text1[j-1] == text2[i-1]:
+                    dp[i][j] = dp[i-1][j-1]+1 
+                else:
+                    dp[i][j] = max(dp[i-1][j], dp[i][j-1])
+        return dp[-1][-1]
+```
+
 
 Go：
 
@@ -142,4 +174,4 @@ Go：
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)
 * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-<div align="center"><img src=../pics/公众号.png width=450 alt=> </img></div>
+<div align="center"><img src=../pics/公众号.png width=450 alt=> </img></div>
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