feat: add solutions to lcof2 problem: No.032

Author: yanglbme

lcof2/剑指 Offer II 032. 有效的变位词/README.md

@@ -46,27 +46,103 @@
 
 <p><meta charset="UTF-8" />注意：本题与主站 242&nbsp;题相似（字母异位词定义不同）：<a href="https://leetcode-cn.com/problems/valid-anagram/">https://leetcode-cn.com/problems/valid-anagram/</a></p>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+数组或哈希表累加 s 中每个字符出现的次数，再减去 t 中对应的每个字符出现的次数。遍历结束后，若字符中出现次数不为 0 的情况，返回 false，否则返回 true。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t) or s == t:
+            return False
+        n = len(s)
+        chars = [0] * 26
+        for i in range(n):
+            chars[ord(s[i]) - ord('a')] += 1
+            chars[ord(t[i]) - ord('a')] -= 1
+        for c in chars:
+            if c != 0:
+                return False
+        return True
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public boolean isAnagram(String s, String t) {
+        int n;
+        if ((n = s.length()) != t.length() || (Objects.equals(s, t))) {
+            return false;
+        }
+        int[] chars = new int[26];
+        for (int i = 0; i < n; ++i) {
+            ++chars[s.charAt(i) - 'a'];
+            --chars[t.charAt(i) - 'a'];
+        }
+        for (int c : chars) {
+            if (c != 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isAnagram(string s, string t) {
+        if (s.size() != t.size() || s == t)
+            return false;
+        vector<int> chars(26, 0);
+        for (int i = 0, n = s.size(); i < n; ++i)
+        {
+            ++chars[s[i] - 'a'];
+            --chars[t[i] - 'a'];
+        }
+        for (int c : chars)
+        {
+            if (c != 0)
+                return false;
+        }
+        return true;
+    }
+};
+```
 
+### **Go**
+
+```go
+func isAnagram(s string, t string) bool {
+    if len(s) != len(t) || s == t {
+        return false
+    }
+    var chars [26]int
+    for i := 0; i < len(s); i++ {
+        chars[s[i]-'a']++
+        chars[t[i]-'a']--
+    }
+    for _, c := range chars {
+        if c != 0 {
+            return false
+        }
+    }
+    return true
+}
 ```
 
 ### **...**

lcof2/剑指 Offer II 032. 有效的变位词/Solution.cpp

@@ -0,0 +1,19 @@
+class Solution {
+public:
+    bool isAnagram(string s, string t) {
+        if (s.size() != t.size() || s == t)
+            return false;
+        vector<int> chars(26, 0);
+        for (int i = 0, n = s.size(); i < n; ++i)
+        {
+            ++chars[s[i] - 'a'];
+            --chars[t[i] - 'a'];
+        }
+        for (int c : chars)
+        {
+            if (c != 0)
+                return false;
+        }
+        return true;
+    }
+};

lcof2/剑指 Offer II 032. 有效的变位词/Solution.go

@@ -0,0 +1,16 @@
+func isAnagram(s string, t string) bool {
+    if len(s) != len(t) || s == t {
+        return false
+    }
+    var chars [26]int
+    for i := 0; i < len(s); i++ {
+        chars[s[i]-'a']++
+        chars[t[i]-'a']--
+    }
+    for _, c := range chars {
+        if c != 0 {
+            return false
+        }
+    }
+    return true
+}

lcof2/剑指 Offer II 032. 有效的变位词/Solution.java

@@ -0,0 +1,19 @@
+class Solution {
+    public boolean isAnagram(String s, String t) {
+        int n;
+        if ((n = s.length()) != t.length() || (Objects.equals(s, t))) {
+            return false;
+        }
+        int[] chars = new int[26];
+        for (int i = 0; i < n; ++i) {
+            ++chars[s.charAt(i) - 'a'];
+            --chars[t.charAt(i) - 'a'];
+        }
+        for (int c : chars) {
+            if (c != 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}

lcof2/剑指 Offer II 032. 有效的变位词/Solution.py

@@ -0,0 +1,13 @@
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t) or s == t:
+            return False
+        n = len(s)
+        chars = [0] * 26
+        for i in range(n):
+            chars[ord(s[i]) - ord('a')] += 1
+            chars[ord(t[i]) - ord('a')] -= 1
+        for c in chars:
+            if c != 0:
+                return False
+        return True

solution/0200-0299/0242.Valid Anagram/README.md

@@ -30,7 +30,7 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-哈希表解决。
+数组或哈希表累加 s 中每个字符出现的次数，再减去 t 中对应的每个字符出现的次数。遍历结束后，若字符中出现次数不为 0 的情况，返回 false，否则返回 true。
 
 <!-- tabs:start -->
 
@@ -48,8 +48,8 @@ class Solution:
         for i in range(n):
             chars[ord(s[i]) - ord('a')] += 1
             chars[ord(t[i]) - ord('a')] -= 1
-        for i in range(26):
-            if chars[i] != 0:
+        for c in chars:
+            if c != 0:
                 return False
         return True
 ```
@@ -70,8 +70,8 @@ class Solution {
             ++chars[s.charAt(i) - 'a'];
             --chars[t.charAt(i) - 'a'];
         }
-        for (int i = 0; i < 26; ++i) {
-            if (chars[i] != 0) {
+        for (int c : chars) {
+            if (c != 0) {
                 return false;
             }
         }
@@ -101,18 +101,18 @@ function isAnagram(s: string, t: string): boolean {
 class Solution {
 public:
     bool isAnagram(string s, string t) {
-        if (s.size() != t.size()) {
+        if (s.size() != t.size())
             return false;
-        }
         vector<int> chars(26, 0);
-        for (int i = 0, n = s.size(); i < n; ++i) {
+        for (int i = 0, n = s.size(); i < n; ++i)
+        {
             ++chars[s[i] - 'a'];
             --chars[t[i] - 'a'];
         }
-        for (int i = 0; i < 26; ++i) {
-            if (chars[i] != 0) {
+        for (int c : chars)
+        {
+            if (c != 0)
                 return false;
-            }
         }
         return true;
     }
@@ -131,8 +131,8 @@ func isAnagram(s string, t string) bool {
 		chars[s[i]-'a']++
 		chars[t[i]-'a']--
 	}
-	for i := 0; i < 26; i++ {
-		if chars[i] != 0 {
+	for _, c := range chars {
+		if c != 0 {
 			return false
 		}
 	}

solution/0200-0299/0242.Valid Anagram/README_EN.md

@@ -42,8 +42,8 @@ class Solution:
         for i in range(n):
             chars[ord(s[i]) - ord('a')] += 1
             chars[ord(t[i]) - ord('a')] -= 1
-        for i in range(26):
-            if chars[i] != 0:
+        for c in chars:
+            if c != 0:
                 return False
         return True
 ```
@@ -62,8 +62,8 @@ class Solution {
             ++chars[s.charAt(i) - 'a'];
             --chars[t.charAt(i) - 'a'];
         }
-        for (int i = 0; i < 26; ++i) {
-            if (chars[i] != 0) {
+        for (int c : chars) {
+            if (c != 0) {
                 return false;
             }
         }
@@ -93,18 +93,18 @@ function isAnagram(s: string, t: string): boolean {
 class Solution {
 public:
     bool isAnagram(string s, string t) {
-        if (s.size() != t.size()) {
+        if (s.size() != t.size())
             return false;
-        }
         vector<int> chars(26, 0);
-        for (int i = 0, n = s.size(); i < n; ++i) {
+        for (int i = 0, n = s.size(); i < n; ++i)
+        {
             ++chars[s[i] - 'a'];
             --chars[t[i] - 'a'];
         }
-        for (int i = 0; i < 26; ++i) {
-            if (chars[i] != 0) {
+        for (int c : chars)
+        {
+            if (c != 0)
                 return false;
-            }
         }
         return true;
     }
@@ -123,8 +123,8 @@ func isAnagram(s string, t string) bool {
 		chars[s[i]-'a']++
 		chars[t[i]-'a']--
 	}
-	for i := 0; i < 26; i++ {
-		if chars[i] != 0 {
+	for _, c := range chars {
+		if c != 0 {
 			return false
 		}
 	}

solution/0200-0299/0242.Valid Anagram/Solution.cpp

@@ -1,18 +1,18 @@
 class Solution {
 public:
     bool isAnagram(string s, string t) {
-        if (s.size() != t.size()) {
+        if (s.size() != t.size())
             return false;
-        }
         vector<int> chars(26, 0);
-        for (int i = 0, n = s.size(); i < n; ++i) {
+        for (int i = 0, n = s.size(); i < n; ++i)
+        {
             ++chars[s[i] - 'a'];
             --chars[t[i] - 'a'];
         }
-        for (int i = 0; i < 26; ++i) {
-            if (chars[i] != 0) {
+        for (int c : chars)
+        {
+            if (c != 0)
                 return false;
-            }
         }
         return true;
     }

solution/0200-0299/0242.Valid Anagram/Solution.go

@@ -7,8 +7,8 @@ func isAnagram(s string, t string) bool {
 		chars[s[i]-'a']++
 		chars[t[i]-'a']--
 	}
-	for i := 0; i < 26; i++ {
-		if chars[i] != 0 {
+	for _, c := range chars {
+		if c != 0 {
 			return false
 		}
 	}

solution/0200-0299/0242.Valid Anagram/Solution.java

@@ -9,8 +9,8 @@ public boolean isAnagram(String s, String t) {
             ++chars[s.charAt(i) - 'a'];
             --chars[t.charAt(i) - 'a'];
         }
-        for (int i = 0; i < 26; ++i) {
-            if (chars[i] != 0) {
+        for (int c : chars) {
+            if (c != 0) {
                 return false;
             }
         }