feat: update solutions to lcof problem: No.15 (#2862)

Author: yanglbme

lcof/面试题15. 二进制中1的个数/README.md

@@ -68,11 +68,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9
 
 ### 方法一：位运算
 
-由于 `n & (n - 1)` 会消除 $n$ 的二进制表示中的最后一个 $1$，因此对 $n$ 重复该操作，直到 $n$ 变成 $0$，此时的操作次数即为 $n$ 的二进制表示中的 $1$ 的个数。
+由于 $n \& (n - 1)$ 可以消除 $n$ 的二进制表示中最右边的 1，因此不断执行 $n \& (n - 1)$，直到 $n = 0$，统计执行次数即可。
 
-或者，我们可以用 `lowbit` 函数来获取 $n$ 的二进制表示中的最后一个 $1$，然后将 $n$ 减去这个 $1$，再重复该操作，直到 $n$ 变成 $0$，此时的操作次数即为 $n$ 的二进制表示中的 $1$ 的个数。`lowbit(x)=x&(-x)`。
-
-时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为输入的整数。
+时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -81,7 +79,11 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9
 ```python
 class Solution:
     def hammingWeight(self, n: int) -> int:
-        return n.bit_count()
+        ans = 0
+        while n:
+            n &= n - 1
+            ans += 1
+        return ans
 ```
 
 #### Java
@@ -92,7 +94,7 @@ public class Solution {
     public int hammingWeight(int n) {
         int ans = 0;
         while (n != 0) {
-            n &= n - 1;
+            n &= (n - 1);
             ++ans;
         }
         return ans;
@@ -108,7 +110,7 @@ public:
     int hammingWeight(uint32_t n) {
         int ans = 0;
         while (n) {
-            n &= n - 1;
+            n &= (n - 1);
             ++ans;
         }
         return ans;
@@ -119,9 +121,9 @@ public:
 #### Go
 
 ```go
-func hammingWeight(num uint32) (ans int) {
-	for num != 0 {
-		num &= num - 1
+func hammingWeight(n uint32) (ans int) {
+	for n != 0 {
+		n &= n - 1
 		ans++
 	}
 	return
@@ -137,36 +139,25 @@ func hammingWeight(num uint32) (ans int) {
  */
 var hammingWeight = function (n) {
     let ans = 0;
-    while (n != 0) {
+    while (n) {
         n &= n - 1;
         ++ans;
     }
     return ans;
 };
 ```
 
-#### C#
-
-```cs
-public class Solution {
-    public int HammingWeight(uint n) {
-        int ans = 0;
-        while (n != 0) {
-            n &= (n - 1);
-            ++ans;
-        }
-        return ans;
-    }
-}
-```
-
 <!-- tabs:end -->
 
 <!-- solution:end -->
 
-<!-- solution:start-->
+<!-- solution:start -->
+
+### 方法二：位运算（lowbit）
+
+根据位运算的性质，我们知道 $n \& (-n)$ 可以得到 $n$ 的二进制表示中最右边的 $1$，因此不断将 $n$ 减去 $n \& (-n)$，直到 $n = 0$，统计执行次数即可。
 
-### 方法二
+时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -177,7 +168,7 @@ class Solution:
     def hammingWeight(self, n: int) -> int:
         ans = 0
         while n:
-            n &= n - 1
+            n -= n & (-n)
             ans += 1
         return ans
 ```
@@ -205,7 +196,7 @@ class Solution {
 public:
     int hammingWeight(uint32_t n) {
         int ans = 0;
-        while (n != 0) {
+        while (n) {
             n -= n & -n;
             ++ans;
         }
@@ -217,9 +208,9 @@ public:
 #### Go
 
 ```go
-func hammingWeight(num uint32) (ans int) {
-	for num != 0 {
-		num -= num & -num
+func hammingWeight(n uint32) (ans int) {
+	for n != 0 {
+		n -= n & -n
 		ans++
 	}
 	return
@@ -230,26 +221,4 @@ func hammingWeight(num uint32) (ans int) {
 
 <!-- solution:end -->
 
-<!-- solution:start-->
-
-### 方法三
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def hammingWeight(self, n: int) -> int:
-        ans = 0
-        while n:
-            n -= n & (-n)
-            ans += 1
-        return ans
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

lcof/面试题15. 二进制中1的个数/Solution.cpp

@@ -1,11 +1,11 @@
-class Solution {
-public:
-    int hammingWeight(uint32_t n) {
-        int ans = 0;
-        while (n) {
-            n &= n - 1;
-            ++ans;
-        }
-        return ans;
-    }
+class Solution {
+public:
+    int hammingWeight(uint32_t n) {
+        int ans = 0;
+        while (n) {
+            n &= (n - 1);
+            ++ans;
+        }
+        return ans;
+    }
 };

lcof/面试题15. 二进制中1的个数/Solution.cs

@@ -1,7 +1,7 @@
-func hammingWeight(num uint32) (ans int) {
-	for num != 0 {
-		num &= num - 1
-		ans++
-	}
-	return
+func hammingWeight(n uint32) (ans int) {
+	for n != 0 {
+		n &= n - 1
+		ans++
+	}
+	return
 }

lcof/面试题15. 二进制中1的个数/Solution.go

@@ -1,11 +1,11 @@
-public class Solution {
-    // you need to treat n as an unsigned value
-    public int hammingWeight(int n) {
-        int ans = 0;
-        while (n != 0) {
-            n &= n - 1;
-            ++ans;
-        }
-        return ans;
-    }
+public class Solution {
+    // you need to treat n as an unsigned value
+    public int hammingWeight(int n) {
+        int ans = 0;
+        while (n != 0) {
+            n &= (n - 1);
+            ++ans;
+        }
+        return ans;
+    }
 }

lcof/面试题15. 二进制中1的个数/Solution.java

@@ -4,7 +4,7 @@
  */
 var hammingWeight = function (n) {
     let ans = 0;
-    while (n != 0) {
+    while (n) {
         n &= n - 1;
         ++ans;
     }

lcof/面试题15. 二进制中1的个数/Solution.js

@@ -1,3 +1,7 @@
-class Solution:
-    def hammingWeight(self, n: int) -> int:
-        return n.bit_count()
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        ans = 0
+        while n:
+            n &= n - 1
+            ans += 1
+        return ans

lcof/面试题15. 二进制中1的个数/Solution.py

@@ -1,11 +1,11 @@
-class Solution {
-public:
-    int hammingWeight(uint32_t n) {
-        int ans = 0;
-        while (n != 0) {
-            n -= n & -n;
-            ++ans;
-        }
-        return ans;
-    }
+class Solution {
+public:
+    int hammingWeight(uint32_t n) {
+        int ans = 0;
+        while (n) {
+            n -= n & -n;
+            ++ans;
+        }
+        return ans;
+    }
 };

lcof/面试题15. 二进制中1的个数/Solution2.cpp

@@ -1,7 +1,7 @@
-func hammingWeight(num uint32) (ans int) {
-	for num != 0 {
-		num -= num & -num
-		ans++
-	}
-	return
+func hammingWeight(n uint32) (ans int) {
+	for n != 0 {
+		n -= n & -n
+		ans++
+	}
+	return
 }

lcof/面试题15. 二进制中1的个数/Solution2.go

@@ -1,11 +1,11 @@
-public class Solution {
-    // you need to treat n as an unsigned value
-    public int hammingWeight(int n) {
-        int ans = 0;
-        while (n != 0) {
-            n -= n & -n;
-            ++ans;
-        }
-        return ans;
-    }
+public class Solution {
+    // you need to treat n as an unsigned value
+    public int hammingWeight(int n) {
+        int ans = 0;
+        while (n != 0) {
+            n -= n & -n;
+            ++ans;
+        }
+        return ans;
+    }
 }

lcof/面试题15. 二进制中1的个数/Solution2.java

@@ -1,7 +1,7 @@
-class Solution:
-    def hammingWeight(self, n: int) -> int:
-        ans = 0
-        while n:
-            n &= n - 1
-            ans += 1
-        return ans
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        ans = 0
+        while n:
+            n -= n & (-n)
+            ans += 1
+        return ans