feat: add solutions to lc problem: No.0392

No.0392.Is Subsequence

Author: yanglbme

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

@@ -50,6 +50,10 @@
 
 **方法一：双指针**
 
+我们定义两个指针 $i$ 和 $j$，分别指向字符串 $s$ 和 $t$ 的初始位置。每次我们比较两个指针指向的字符，如果相同，则两个指针同时右移；如果不同，则只有 $j$ 右移。当指针 $i$ 移动到字符串 $s$ 的末尾时，说明 $s$ 是 $t$ 的子序列。
+
+时间复杂度 $O(m + n)$，其中 $m$ 和 $n$ 分别是字符串 $s$ 和 $t$ 的长度。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -59,12 +63,12 @@
 ```python
 class Solution:
     def isSubsequence(self, s: str, t: str) -> bool:
-        i, j, m, n = 0, 0, len(s), len(t)
-        while i < m and j < n:
+        i = j = 0
+        while i < len(s) and j < len(t):
             if s[i] == t[j]:
                 i += 1
             j += 1
-        return i == m
+        return i == len(s)
 ```
 
 ### **Java**
@@ -95,9 +99,10 @@ public:
     bool isSubsequence(string s, string t) {
         int m = s.size(), n = t.size();
         int i = 0, j = 0;
-        while (i < m && j < n) {
-            if (s[i] == t[j]) ++i;
-            ++j;
+        for (; i < m && j < n; ++j) {
+            if (s[i] == t[j]) {
+                ++i;
+            }
         }
         return i == m;
     }
@@ -123,15 +128,15 @@ func isSubsequence(s string, t string) bool {
 
 ```ts
 function isSubsequence(s: string, t: string): boolean {
-    let m = s.length,
-        n = t.length;
+    const m = s.length;
+    const n = t.length;
     let i = 0;
-    for (let j = 0; j < n && i < m; ++j) {
-        if (s.charAt(i) == t.charAt(j)) {
+    for (let j = 0; i < m && j < n; ++j) {
+        if (s[i] === t[j]) {
             ++i;
         }
     }
-    return i == m;
+    return i === m;
 }
 ```
 
@@ -160,19 +165,33 @@ impl Solution {
 ### **C**
 
 ```c
-bool isSubsequence(char *s, char *t) {
-    int n = strlen(s);
+bool isSubsequence(char * s, char * t){
+    int m = strlen(s);
+    int n = strlen(t);
     int i = 0;
-    for (int j = 0; j < n; j++) {
-        while (t[i] && t[i] != s[j]) {
-            i++;
+    for (int j = 0; i < m && j < n; ++j) {
+        if (s[i] == t[j]) {
+            ++i;
         }
-        if (!t[i]) {
-            return 0;
+    }
+    return i == m;
+}
+```
+
+### **C#**
+
+```cs
+public class Solution {
+    public bool IsSubsequence(string s, string t) {
+        int m = s.Length, n = t.Length;
+        int i = 0, j = 0;
+        for (; i < m && j < n; ++j) {
+            if (s[i] == t[j]) {
+                ++i;
+            }
         }
-        i++;
+        return i == m;
     }
-    return 1;
 }
 ```
 

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

@@ -30,19 +30,25 @@
 
 ## Solutions
 
+**Approach 1: Two Pointers**
+
+We define two pointers $i$ and $j$ to point to the initial position of the string $s$ and $t$ respectively. Each time we compare the two characters pointed to by the two pointers, if they are the same, both pointers move right at the same time; if they are not the same, only $j$ moves right. When the pointer $i$ moves to the end of the string $s$, it means that $s$ is the subsequence of $t$.
+
+The time complexity is $O(m + n)$, where $m$ and $n$ are the lengths of the strings $s$ and $t$ respectively. The space complexity is $O(1)$.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
 class Solution:
     def isSubsequence(self, s: str, t: str) -> bool:
-        i, j, m, n = 0, 0, len(s), len(t)
-        while i < m and j < n:
+        i = j = 0
+        while i < len(s) and j < len(t):
             if s[i] == t[j]:
                 i += 1
             j += 1
-        return i == m
+        return i == len(s)
 ```
 
 ### **Java**
@@ -71,9 +77,10 @@ public:
     bool isSubsequence(string s, string t) {
         int m = s.size(), n = t.size();
         int i = 0, j = 0;
-        while (i < m && j < n) {
-            if (s[i] == t[j]) ++i;
-            ++j;
+        for (; i < m && j < n; ++j) {
+            if (s[i] == t[j]) {
+                ++i;
+            }
         }
         return i == m;
     }
@@ -99,15 +106,15 @@ func isSubsequence(s string, t string) bool {
 
 ```ts
 function isSubsequence(s: string, t: string): boolean {
-    let m = s.length,
-        n = t.length;
+    const m = s.length;
+    const n = t.length;
     let i = 0;
-    for (let j = 0; j < n && i < m; ++j) {
-        if (s.charAt(i) == t.charAt(j)) {
+    for (let j = 0; i < m && j < n; ++j) {
+        if (s[i] === t[j]) {
             ++i;
         }
     }
-    return i == m;
+    return i === m;
 }
 ```
 
@@ -136,19 +143,33 @@ impl Solution {
 ### **C**
 
 ```c
-bool isSubsequence(char *s, char *t) {
-    int n = strlen(s);
+bool isSubsequence(char * s, char * t){
+    int m = strlen(s);
+    int n = strlen(t);
     int i = 0;
-    for (int j = 0; j < n; j++) {
-        while (t[i] && t[i] != s[j]) {
-            i++;
+    for (int j = 0; i < m && j < n; ++j) {
+        if (s[i] == t[j]) {
+            ++i;
         }
-        if (!t[i]) {
-            return 0;
+    }
+    return i == m;
+}
+```
+
+### **C#**
+
+```cs
+public class Solution {
+    public bool IsSubsequence(string s, string t) {
+        int m = s.Length, n = t.Length;
+        int i = 0, j = 0;
+        for (; i < m && j < n; ++j) {
+            if (s[i] == t[j]) {
+                ++i;
+            }
         }
-        i++;
+        return i == m;
     }
-    return 1;
 }
 ```
 

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

@@ -1,14 +1,11 @@
-bool isSubsequence(char* s, char* t) {
-    int n = strlen(s);
+bool isSubsequence(char * s, char * t){
+    int m = strlen(s);
+    int n = strlen(t);
     int i = 0;
-    for (int j = 0; j < n; j++) {
-        while (t[i] && t[i] != s[j]) {
-            i++;
+    for (int j = 0; i < m && j < n; ++j) {
+        if (s[i] == t[j]) {
+            ++i;
         }
-        if (!t[i]) {
-            return 0;
-        }
-        i++;
     }
-    return 1;
-}
+    return i == m;
+}

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

@@ -3,9 +3,10 @@ class Solution {
     bool isSubsequence(string s, string t) {
         int m = s.size(), n = t.size();
         int i = 0, j = 0;
-        while (i < m && j < n) {
-            if (s[i] == t[j]) ++i;
-            ++j;
+        for (; i < m && j < n; ++j) {
+            if (s[i] == t[j]) {
+                ++i;
+            }
         }
         return i == m;
     }

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

@@ -0,0 +1,12 @@
+public class Solution {
+    public bool IsSubsequence(string s, string t) {
+        int m = s.Length, n = t.Length;
+        int i = 0, j = 0;
+        for (; i < m && j < n; ++j) {
+            if (s[i] == t[j]) {
+                ++i;
+            }
+        }
+        return i == m;
+    }
+}

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

@@ -1,8 +1,8 @@
 class Solution:
     def isSubsequence(self, s: str, t: str) -> bool:
-        i, j, m, n = 0, 0, len(s), len(t)
-        while i < m and j < n:
+        i = j = 0
+        while i < len(s) and j < len(t):
             if s[i] == t[j]:
                 i += 1
             j += 1
-        return i == m
+        return i == len(s)

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

@@ -1,11 +1,11 @@
 function isSubsequence(s: string, t: string): boolean {
-    let m = s.length,
-        n = t.length;
+    const m = s.length;
+    const n = t.length;
     let i = 0;
-    for (let j = 0; j < n && i < m; ++j) {
-        if (s.charAt(i) == t.charAt(j)) {
+    for (let j = 0; i < m && j < n; ++j) {
+        if (s[i] === t[j]) {
             ++i;
         }
     }
-    return i == m;
+    return i === m;
 }