feat: add solutions to lcci problem: No.05.06

No.05.06.Convert Integer

Author: yanglbme

lcci/05.06.Convert Integer/README.md

@@ -31,15 +31,24 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：位运算**
+
+我们将 A 和 B 进行异或运算，得到的结果中 $1$ 的个数即为需要改变的位数。
+
+时间复杂度 $O(\log n)$，其中 $n$ 为 A 和 B 的最大值。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
-
+class Solution:
+    def convertInteger(self, A: int, B: int) -> int:
+        A &= 0xFFFFFFFF
+        B &= 0xFFFFFFFF
+        return (A ^ B).bit_count()
 ```
 
 ### **Java**
@@ -54,6 +63,26 @@ class Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int convertInteger(int A, int B) {
+        unsigned int c = A ^ B;
+        return __builtin_popcount(c);
+    }
+};
+```
+
+### **Go**
+
+```go
+func convertInteger(A int, B int) int {
+	return bits.OnesCount32(uint32(A ^ B))
+}
+```
+
 ### **TypeScript**
 
 ```ts

lcci/05.06.Convert Integer/README_EN.md

@@ -51,8 +51,11 @@
 ### **Python3**
 
 ```python
-
-
+class Solution:
+    def convertInteger(self, A: int, B: int) -> int:
+        A &= 0xFFFFFFFF
+        B &= 0xFFFFFFFF
+        return (A ^ B).bit_count()
 ```
 
 ### **Java**
@@ -65,6 +68,26 @@ class Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int convertInteger(int A, int B) {
+        unsigned int c = A ^ B;
+        return __builtin_popcount(c);
+    }
+};
+```
+
+### **Go**
+
+```go
+func convertInteger(A int, B int) int {
+	return bits.OnesCount32(uint32(A ^ B))
+}
+```
+
 ### **TypeScript**
 
 ```ts

lcci/05.06.Convert Integer/Solution.cpp

@@ -0,0 +1,7 @@
+class Solution {
+public:
+    int convertInteger(int A, int B) {
+        unsigned int c = A ^ B;
+        return __builtin_popcount(c);
+    }
+};

lcci/05.06.Convert Integer/Solution.go

@@ -0,0 +1,3 @@
+func convertInteger(A int, B int) int {
+	return bits.OnesCount32(uint32(A ^ B))
+}

lcci/05.06.Convert Integer/Solution.py

@@ -0,0 +1,5 @@
+class Solution:
+    def convertInteger(self, A: int, B: int) -> int:
+        A &= 0xFFFFFFFF
+        B &= 0xFFFFFFFF
+        return (A ^ B).bit_count()