diff --git a/problems/3.longest-substring-without-repeating-characters.md b/problems/3.longest-substring-without-repeating-characters.md
index a36daedb9..b14da9627 100644
--- a/problems/3.longest-substring-without-repeating-characters.md
+++ b/problems/3.longest-substring-without-repeating-characters.md
@@ -49,7 +49,7 @@ https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
 
 2. 如果当前遍历到的字符出现过，则缩小窗口（左边索引向右移动），然后继续观察当前遍历到的字符；
 
-3. 重复（1）（2），直到左边索引无法再移动；
+3. 重复（1）（2），直到窗口内无重复元素；
 
 4. 维护一个全局最大窗口 res，每次用出现过的窗口大小来更新结果 res，最后返回 res 获取结果;
 
@@ -66,7 +66,65 @@ https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
 
 ## 代码
 
-代码支持：Python3
+代码支持：C++，Java，Python3
+
+
+C++ Code:
+
+```c++
+class Solution {
+public:
+    int lengthOfLongestSubstring(string s) {
+        
+        int ans = 0, start = 0;
+        int n = s.length();
+        // 
+        map<char, int> mp;
+
+        for(int i=0;i<n;i++)
+        {
+            char alpha = s[i];
+            if(mp.count(alpha))
+            {
+                start = max(start, mp[alpha]+1);
+            }
+            ans = max(ans, i-start+1);
+            // 字符位置
+            mp[alpha] = i;
+        }
+
+        return ans;
+    }
+};
+```
+
+
+Java Code:
+
+```java
+class Solution {
+    public int lengthOfLongestSubstring(String s) {
+        int ans = 0, start = 0;
+        int n = s.length();
+        // 
+        Map<Character, Integer> map = new HashMap<>();
+
+        for(int i=0;i<n;i++)
+        {
+            char alpha = s.charAt(i);
+            if(map.containsKey(alpha))
+            {
+                start = Math.max(start, map.get(alpha)+1);
+            }
+            ans = Math.max(ans, i-start+1);
+            // 字符位置
+            map.put(alpha, i);
+        }
+
+        return ans;
+    }
+}
+```
 
 Python3 Code: