feat: add solutions to lc problem: No.0869

No.0869.Reordered Power of 2

Author: yanglbme

solution/0800-0899/0869.Reordered Power of 2/README.md

@@ -53,7 +53,6 @@
 	<li><code>1 &lt;= N &lt;= 10^9</code></li>
 </ol>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
@@ -65,15 +64,95 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def reorderedPowerOf2(self, n: int) -> bool:
+        def convert(n):
+            counter = [0] * 10
+            while n > 0:
+                counter[n % 10] += 1
+                n //= 10
+            return counter
+
+        i, s = 1, convert(n)
+        while i <= 10 ** 9:
+            if convert(i) == s:
+                return True
+            i <<= 1
+        return False
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public boolean reorderedPowerOf2(int n) {
+        String s = convert(n);
+        for (int i = 1; i <= Math.pow(10, 9); i <<= 1) {
+            if (s.equals(convert(i))) {
+                return true;
+            }
+        }
+        return false;
+    }
+
+    private String convert(int n) {
+        char[] counter = new char[10];
+        while (n > 0) {
+            ++counter[n % 10];
+            n /= 10;
+        }
+        return new String(counter);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool reorderedPowerOf2(int n) {
+        vector<int> s = convert(n);
+        for (int i = 1; i <= pow(10, 9); i <<= 1)
+            if (s == convert(i)) return true;
+        return false;
+    }
+
+    vector<int> convert(int n) {
+        vector<int> counter(10);
+        while (n)
+        {
+            ++counter[n % 10];
+            n /= 10;
+        }
+        return counter;
+    }
+};
+```
 
+### **Go**
+
+```go
+func reorderedPowerOf2(n int) bool {
+	convert := func(n int) []byte {
+		counter := make([]byte, 10)
+		for n > 0 {
+			counter[n%10]++
+			n /= 10
+		}
+		return counter
+	}
+
+	s := convert(n)
+	for i := 1; i <= 1e9; i <<= 1 {
+		if bytes.Equal(s, convert(i)) {
+			return true
+		}
+	}
+	return false
+}
 ```
 
 ### **...**

solution/0800-0899/0869.Reordered Power of 2/README_EN.md

@@ -51,21 +51,100 @@
 	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def reorderedPowerOf2(self, n: int) -> bool:
+        def convert(n):
+            counter = [0] * 10
+            while n > 0:
+                counter[n % 10] += 1
+                n //= 10
+            return counter
+
+        i, s = 1, convert(n)
+        while i <= 10 ** 9:
+            if convert(i) == s:
+                return True
+            i <<= 1
+        return False
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public boolean reorderedPowerOf2(int n) {
+        String s = convert(n);
+        for (int i = 1; i <= Math.pow(10, 9); i <<= 1) {
+            if (s.equals(convert(i))) {
+                return true;
+            }
+        }
+        return false;
+    }
+
+    private String convert(int n) {
+        char[] counter = new char[10];
+        while (n > 0) {
+            ++counter[n % 10];
+            n /= 10;
+        }
+        return new String(counter);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool reorderedPowerOf2(int n) {
+        vector<int> s = convert(n);
+        for (int i = 1; i <= pow(10, 9); i <<= 1)
+            if (s == convert(i)) return true;
+        return false;
+    }
+
+    vector<int> convert(int n) {
+        vector<int> counter(10);
+        while (n)
+        {
+            ++counter[n % 10];
+            n /= 10;
+        }
+        return counter;
+    }
+};
+```
 
+### **Go**
+
+```go
+func reorderedPowerOf2(n int) bool {
+	convert := func(n int) []byte {
+		counter := make([]byte, 10)
+		for n > 0 {
+			counter[n%10]++
+			n /= 10
+		}
+		return counter
+	}
+
+	s := convert(n)
+	for i := 1; i <= 1e9; i <<= 1 {
+		if bytes.Equal(s, convert(i)) {
+			return true
+		}
+	}
+	return false
+}
 ```
 
 ### **...**

solution/0800-0899/0869.Reordered Power of 2/Solution.cpp

@@ -0,0 +1,19 @@
+class Solution {
+public:
+    bool reorderedPowerOf2(int n) {
+        vector<int> s = convert(n);
+        for (int i = 1; i <= pow(10, 9); i <<= 1)
+            if (s == convert(i)) return true;
+        return false;
+    }
+
+    vector<int> convert(int n) {
+        vector<int> counter(10);
+        while (n) 
+        {
+            ++counter[n % 10];
+            n /= 10;
+        }
+        return counter;
+    }
+};

solution/0800-0899/0869.Reordered Power of 2/Solution.go

@@ -0,0 +1,18 @@
+func reorderedPowerOf2(n int) bool {
+	convert := func(n int) []byte {
+		counter := make([]byte, 10)
+		for n > 0 {
+			counter[n%10]++
+			n /= 10
+		}
+		return counter
+	}
+
+	s := convert(n)
+	for i := 1; i <= 1e9; i <<= 1 {
+		if bytes.Equal(s, convert(i)) {
+			return true
+		}
+	}
+	return false
+}

solution/0800-0899/0869.Reordered Power of 2/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public boolean reorderedPowerOf2(int n) {
+        String s = convert(n);
+        for (int i = 1; i <= Math.pow(10, 9); i <<= 1) {
+            if (s.equals(convert(i))) {
+                return true;
+            }
+        }
+        return false;
+    }
+
+    private String convert(int n) {
+        char[] counter = new char[10];
+        while (n > 0) {
+            ++counter[n % 10];
+            n /= 10;
+        }
+        return new String(counter);
+    }
+}

solution/0800-0899/0869.Reordered Power of 2/Solution.py

@@ -0,0 +1,15 @@
+class Solution:
+    def reorderedPowerOf2(self, n: int) -> bool:
+        def convert(n):
+            counter = [0] * 10
+            while n > 0:
+                counter[n % 10] += 1
+                n //= 10
+            return counter
+
+        i, s = 1, convert(n)
+        while i <= 10 ** 9:
+            if convert(i) == s:
+                return True
+            i <<= 1
+        return False