feat: add python and java solutions to leetcode problem: No.0392

Author: yanglbme

solution/0300-0399/0392.Is Subsequence/README.md

@@ -33,22 +33,44 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+双指针遍历即可。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        m, n = len(s), len(t)
+        i = j = 0
+        while i < m and j < n:
+            if s[i] == t[j]:
+                i += 1
+            j += 1
+        return i == m
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public boolean isSubsequence(String s, String t) {
+        int m = s.length(), n = t.length();
+        int i = 0, j = 0;
+        while (i < m && j < n) {
+            if (s.charAt(i) == t.charAt(j)) {
+                ++i;
+            }
+            ++j;
+        }
+        return i == m;
+    }
+}
 ```
 
 ### **...**

solution/0300-0399/0392.Is Subsequence/README_EN.md

@@ -59,13 +59,33 @@ If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you wan
 ### **Python3**
 
 ```python
-
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        m, n = len(s), len(t)
+        i = j = 0
+        while i < m and j < n:
+            if s[i] == t[j]:
+                i += 1
+            j += 1
+        return i == m
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public boolean isSubsequence(String s, String t) {
+        int m = s.length(), n = t.length();
+        int i = 0, j = 0;
+        while (i < m && j < n) {
+            if (s.charAt(i) == t.charAt(j)) {
+                ++i;
+            }
+            ++j;
+        }
+        return i == m;
+    }
+}
 ```
 
 ### **...**

solution/0300-0399/0392.Is Subsequence/Solution.java

@@ -1,12 +1,13 @@
 class Solution {
     public boolean isSubsequence(String s, String t) {
-        int index = -1;
-        for (char c : s.toCharArray()) {
-            index = t.indexOf(c, index + 1);
-            if (index == -1) {
-                return false;
+        int m = s.length(), n = t.length();
+        int i = 0, j = 0;
+        while (i < m && j < n) {
+            if (s.charAt(i) == t.charAt(j)) {
+                ++i;
             }
+            ++j;
         }
-        return true;
+        return i == m;
     }
-}
+}

solution/0300-0399/0392.Is Subsequence/Solution.py

@@ -0,0 +1,9 @@
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        m, n = len(s), len(t)
+        i = j = 0
+        while i < m and j < n:
+            if s[i] == t[j]:
+                i += 1
+            j += 1
+        return i == m