feat: add python and java solutions to leetcode problem: No.0205

Author: yanglbme

solution/0200-0299/0205.Isomorphic Strings/README.md

@@ -41,15 +41,38 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        a2b, b2a = {}, {}
+        n = len(s)
+        for i in range(n):
+            a, b = s[i], t[i]
+            if (a in a2b and a2b[a] != b) or (b in b2a and b2a[b] != a):
+                return False
+            a2b[a] = b
+            b2a[b] = a
+        return True
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public boolean isIsomorphic(String s, String t) {
+        int n = s.length();
+        Map<Character, Character> a2b = new HashMap<>();
+        Map<Character, Character> b2a = new HashMap<>();
+        for (int i = 0; i < n; ++i) {
+            char a = s.charAt(i), b = t.charAt(i);
+            if ((a2b.containsKey(a) && a2b.get(a) != b) || (b2a.containsKey(b) && b2a.get(b) != a)) return false;
+            a2b.put(a, b);
+            b2a.put(b, a);
+        }
+        return true;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0205.Isomorphic Strings/README_EN.md

@@ -47,13 +47,36 @@ You may assume both <b><i>s&nbsp;</i></b>and <b><i>t&nbsp;</i></b>have the same
 ### **Python3**
 
 ```python
-
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        a2b, b2a = {}, {}
+        n = len(s)
+        for i in range(n):
+            a, b = s[i], t[i]
+            if (a in a2b and a2b[a] != b) or (b in b2a and b2a[b] != a):
+                return False
+            a2b[a] = b
+            b2a[b] = a
+        return True
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public boolean isIsomorphic(String s, String t) {
+        int n = s.length();
+        Map<Character, Character> a2b = new HashMap<>();
+        Map<Character, Character> b2a = new HashMap<>();
+        for (int i = 0; i < n; ++i) {
+            char a = s.charAt(i), b = t.charAt(i);
+            if ((a2b.containsKey(a) && a2b.get(a) != b) || (b2a.containsKey(b) && b2a.get(b) != a)) return false;
+            a2b.put(a, b);
+            b2a.put(b, a);
+        }
+        return true;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0205.Isomorphic Strings/Solution.java

@@ -1,36 +1,14 @@
-public class Solution {
+class Solution {
     public boolean isIsomorphic(String s, String t) {
-
-		if (s == null || t == null) {
-			return false;
-		}
-
-		if (s.length() != t.length()) {
-			return false;
-		}
-
-		Map<Character, Character> map = new HashMap<Character, Character>();
-		Set<Character> set = new HashSet<Character>();
-
-		for (int i = 0; i < s.length(); i++) {
-			char c1 = s.charAt(i);
-			char c2 = t.charAt(i);
-
-			if (map.containsKey(c1)) {
-				if (map.get(c1) != c2) {
-					return false;
-				}
-			} else {
-				if (set.contains(c2)) {
-					return false;
-				}
-				map.put(c1, c2);
-				set.add(c2);
-			}
-		}
-
-		return true;
-	
-        
+        int n = s.length();
+        Map<Character, Character> a2b = new HashMap<>();
+        Map<Character, Character> b2a = new HashMap<>();
+        for (int i = 0; i < n; ++i) {
+            char a = s.charAt(i), b = t.charAt(i);
+            if ((a2b.containsKey(a) && a2b.get(a) != b) || (b2a.containsKey(b) && b2a.get(b) != a)) return false;
+            a2b.put(a, b);
+            b2a.put(b, a);
+        }
+        return true;
     }
 }

solution/0200-0299/0205.Isomorphic Strings/Solution.py

@@ -0,0 +1,11 @@
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        a2b, b2a = {}, {}
+        n = len(s)
+        for i in range(n):
+            a, b = s[i], t[i]
+            if (a in a2b and a2b[a] != b) or (b in b2a and b2a[b] != a):
+                return False
+            a2b[a] = b
+            b2a[b] = a
+        return True