fix: Use int instead of float for the example code of log time comple…

…xity (#1164) * Use int instead of float for the example code of log time complexity * Bug fixes * Bug fixes
krahets · Mar 22, 2024 · 3ea91bd · 3ea91bd
1 parent fc8473c
commit 3ea91bd
Show file tree

Hide file tree

Showing 12 changed files with 69 additions and 69 deletions.
diff --git a/codes/c/chapter_computational_complexity/time_complexity.c b/codes/c/chapter_computational_complexity/time_complexity.c
@@ -90,7 +90,7 @@ int expRecur(int n) {
 }
 
 /* 对数阶（循环实现） */
-int logarithmic(float n) {
+int logarithmic(int n) {
     int count = 0;
     while (n > 1) {
         n = n / 2;
@@ -100,14 +100,14 @@ int logarithmic(float n) {
 }
 
 /* 对数阶（递归实现） */
-int logRecur(float n) {
+int logRecur(int n) {
     if (n <= 1)
         return 0;
     return logRecur(n / 2) + 1;
 }
 
 /* 线性对数阶 */
-int linearLogRecur(float n) {
+int linearLogRecur(int n) {
     if (n <= 1)
         return 1;
     int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);

diff --git a/codes/cpp/chapter_computational_complexity/time_complexity.cpp b/codes/cpp/chapter_computational_complexity/time_complexity.cpp
@@ -86,7 +86,7 @@ int expRecur(int n) {
 }
 
 /* 对数阶（循环实现） */
-int logarithmic(float n) {
+int logarithmic(int n) {
     int count = 0;
     while (n > 1) {
         n = n / 2;
@@ -96,14 +96,14 @@ int logarithmic(float n) {
 }
 
 /* 对数阶（递归实现） */
-int logRecur(float n) {
+int logRecur(int n) {
     if (n <= 1)
         return 0;
     return logRecur(n / 2) + 1;
 }
 
 /* 线性对数阶 */
-int linearLogRecur(float n) {
+int linearLogRecur(int n) {
     if (n <= 1)
         return 1;
     int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
@@ -153,12 +153,12 @@ int main() {
     count = expRecur(n);
     cout << "指数阶（递归实现）的操作数量 = " << count << endl;
 
-    count = logarithmic((float)n);
+    count = logarithmic(n);
     cout << "对数阶（循环实现）的操作数量 = " << count << endl;
-    count = logRecur((float)n);
+    count = logRecur(n);
     cout << "对数阶（递归实现）的操作数量 = " << count << endl;
 
-    count = linearLogRecur((float)n);
+    count = linearLogRecur(n);
     cout << "线性对数阶（递归实现）的操作数量 = " << count << endl;
 
     count = factorialRecur(n);

diff --git a/codes/csharp/chapter_computational_complexity/time_complexity.cs b/codes/csharp/chapter_computational_complexity/time_complexity.cs
@@ -10,7 +10,7 @@ public class time_complexity {
     void Algorithm(int n) {
         int a = 1;  // +0（技巧 1）
         a += n;  // +0（技巧 1）
-                    // +n（技巧 2）
+        // +n（技巧 2）
         for (int i = 0; i < 5 * n + 1; i++) {
             Console.WriteLine(0);
         }
@@ -118,7 +118,7 @@ public class time_complexity {
     }
 
     /* 对数阶（循环实现） */
-    int Logarithmic(float n) {
+    int Logarithmic(int n) {
         int count = 0;
         while (n > 1) {
             n /= 2;
@@ -128,13 +128,13 @@ public class time_complexity {
     }
 
     /* 对数阶（递归实现） */
-    int LogRecur(float n) {
+    int LogRecur(int n) {
         if (n <= 1) return 0;
         return LogRecur(n / 2) + 1;
     }
 
     /* 线性对数阶 */
-    int LinearLogRecur(float n) {
+    int LinearLogRecur(int n) {
         if (n <= 1) return 1;
         int count = LinearLogRecur(n / 2) + LinearLogRecur(n / 2);
         for (int i = 0; i < n; i++) {
@@ -181,12 +181,12 @@ public class time_complexity {
         count = ExpRecur(n);
         Console.WriteLine("指数阶（递归实现）的操作数量 = " + count);
 
-        count = Logarithmic((float)n);
+        count = Logarithmic(n);
         Console.WriteLine("对数阶（循环实现）的操作数量 = " + count);
-        count = LogRecur((float)n);
+        count = LogRecur(n);
         Console.WriteLine("对数阶（递归实现）的操作数量 = " + count);
 
-        count = LinearLogRecur((float)n);
+        count = LinearLogRecur(n);
         Console.WriteLine("线性对数阶（递归实现）的操作数量 = " + count);
 
         count = FactorialRecur(n);

diff --git a/codes/dart/chapter_computational_complexity/time_complexity.dart b/codes/dart/chapter_computational_complexity/time_complexity.dart
@@ -87,25 +87,25 @@ int expRecur(int n) {
 }
 
 /* 对数阶（循环实现） */
-int logarithmic(num n) {
+int logarithmic(int n) {
   int count = 0;
   while (n > 1) {
-    n = n / 2;
+    n = n ~/ 2;
     count++;
   }
   return count;
 }
 
 /* 对数阶（递归实现） */
-int logRecur(num n) {
+int logRecur(int n) {
   if (n <= 1) return 0;
-  return logRecur(n / 2) + 1;
+  return logRecur(n ~/ 2) + 1;
 }
 
 /* 线性对数阶 */
-int linearLogRecur(num n) {
+int linearLogRecur(int n) {
   if (n <= 1) return 1;
-  int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
+  int count = linearLogRecur(n ~/ 2) + linearLogRecur(n ~/ 2);
   for (var i = 0; i < n; i++) {
     count++;
   }

diff --git a/codes/go/chapter_computational_complexity/time_complexity.go b/codes/go/chapter_computational_complexity/time_complexity.go
@@ -87,7 +87,7 @@ func expRecur(n int) int {
 }
 
 /* 对数阶（循环实现）*/
-func logarithmic(n float64) int {
+func logarithmic(n int) int {
 	count := 0
 	for n > 1 {
 		n = n / 2
@@ -97,20 +97,20 @@ func logarithmic(n float64) int {
 }
 
 /* 对数阶（递归实现）*/
-func logRecur(n float64) int {
+func logRecur(n int) int {
 	if n <= 1 {
 		return 0
 	}
 	return logRecur(n/2) + 1
 }
 
 /* 线性对数阶 */
-func linearLogRecur(n float64) int {
+func linearLogRecur(n int) int {
 	if n <= 1 {
 		return 1
 	}
 	count := linearLogRecur(n/2) + linearLogRecur(n/2)
-	for i := 0.0; i < n; i++ {
+	for i := 0; i < n; i++ {
 		count++
 	}
 	return count

diff --git a/codes/go/chapter_computational_complexity/time_complexity_test.go b/codes/go/chapter_computational_complexity/time_complexity_test.go
@@ -35,12 +35,12 @@ func TestTimeComplexity(t *testing.T) {
 	count = expRecur(n)
 	fmt.Println("指数阶（递归实现）的操作数量 =", count)
 
-	count = logarithmic(float64(n))
+	count = logarithmic(n)
 	fmt.Println("对数阶（循环实现）的操作数量 =", count)
-	count = logRecur(float64(n))
+	count = logRecur(n)
 	fmt.Println("对数阶（递归实现）的操作数量 =", count)
 
-	count = linearLogRecur(float64(n))
+	count = linearLogRecur(n)
 	fmt.Println("线性对数阶（递归实现）的操作数量 =", count)
 
 	count = factorialRecur(n)

diff --git a/codes/java/chapter_computational_complexity/time_complexity.java b/codes/java/chapter_computational_complexity/time_complexity.java
@@ -87,7 +87,7 @@ static int expRecur(int n) {
     }
 
     /* 对数阶（循环实现） */
-    static int logarithmic(float n) {
+    static int logarithmic(int n) {
         int count = 0;
         while (n > 1) {
             n = n / 2;
@@ -97,14 +97,14 @@ static int logarithmic(float n) {
     }
 
     /* 对数阶（递归实现） */
-    static int logRecur(float n) {
+    static int logRecur(int n) {
         if (n <= 1)
             return 0;
         return logRecur(n / 2) + 1;
     }
 
     /* 线性对数阶 */
-    static int linearLogRecur(float n) {
+    static int linearLogRecur(int n) {
         if (n <= 1)
             return 1;
         int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
@@ -153,12 +153,12 @@ public static void main(String[] args) {
         count = expRecur(n);
         System.out.println("指数阶（递归实现）的操作数量 = " + count);
 
-        count = logarithmic((float) n);
+        count = logarithmic(n);
         System.out.println("对数阶（循环实现）的操作数量 = " + count);
-        count = logRecur((float) n);
+        count = logRecur(n);
         System.out.println("对数阶（递归实现）的操作数量 = " + count);
 
-        count = linearLogRecur((float) n);
+        count = linearLogRecur(n);
         System.out.println("线性对数阶（递归实现）的操作数量 = " + count);
 
         count = factorialRecur(n);

diff --git a/codes/kotlin/chapter_computational_complexity/time_complexity.kt b/codes/kotlin/chapter_computational_complexity/time_complexity.kt
@@ -87,7 +87,7 @@ fun expRecur(n: Int): Int {
 }
 
 /* 对数阶（循环实现） */
-fun logarithmic(n: Float): Int {
+fun logarithmic(n: Int): Int {
     var n1 = n
     var count = 0
     while (n1 > 1) {
@@ -98,14 +98,14 @@ fun logarithmic(n: Float): Int {
 }
 
 /* 对数阶（递归实现） */
-fun logRecur(n: Float): Int {
+fun logRecur(n: Int): Int {
     if (n <= 1)
         return 0
     return logRecur(n / 2) + 1
 }
 
 /* 线性对数阶 */
-fun linearLogRecur(n: Float): Int {
+fun linearLogRecur(n: Int): Int {
     if (n <= 1)
         return 1
     var count = linearLogRecur(n / 2) + linearLogRecur(n / 2)
@@ -153,12 +153,12 @@ fun main() {
     count = expRecur(n)
     println("指数阶（递归实现）的操作数量 = $count")
 
-    count = logarithmic(n.toFloat())
+    count = logarithmic(n)
     println("对数阶（循环实现）的操作数量 = $count")
-    count = logRecur(n.toFloat())
+    count = logRecur(n)
     println("对数阶（递归实现）的操作数量 = $count")
 
-    count = linearLogRecur(n.toFloat())
+    count = linearLogRecur(n)
     println("线性对数阶（递归实现）的操作数量 = $count")
 
     count = factorialRecur(n)

diff --git a/codes/python/chapter_computational_complexity/time_complexity.py b/codes/python/chapter_computational_complexity/time_complexity.py
@@ -77,7 +77,7 @@ def exp_recur(n: int) -> int:
     return exp_recur(n - 1) + exp_recur(n - 1) + 1
 
 
-def logarithmic(n: float) -> int:
+def logarithmic(n: int) -> int:
     """对数阶（循环实现）"""
     count = 0
     while n > 1:
@@ -86,14 +86,14 @@ def logarithmic(n: float) -> int:
     return count
 
 
-def log_recur(n: float) -> int:
+def log_recur(n: int) -> int:
     """对数阶（递归实现）"""
     if n <= 1:
         return 0
     return log_recur(n / 2) + 1
 
 
-def linear_log_recur(n: float) -> int:
+def linear_log_recur(n: int) -> int:
     """线性对数阶"""
     if n <= 1:
         return 1

diff --git a/codes/rust/chapter_computational_complexity/time_complexity.rs b/codes/rust/chapter_computational_complexity/time_complexity.rs
@@ -90,29 +90,29 @@ fn exp_recur(n: i32) -> i32 {
 }
 
 /* 对数阶（循环实现） */
-fn logarithmic(mut n: f32) -> i32 {
+fn logarithmic(mut n: i32) -> i32 {
     let mut count = 0;
-    while n > 1.0 {
-        n = n / 2.0;
+    while n > 1 {
+        n = n / 2;
         count += 1;
     }
     count
 }
 
 /* 对数阶（递归实现） */
-fn log_recur(n: f32) -> i32 {
-    if n <= 1.0 {
+fn log_recur(n: i32) -> i32 {
+    if n <= 1 {
         return 0;
     }
-    log_recur(n / 2.0) + 1
+    log_recur(n / 2) + 1
 }
 
 /* 线性对数阶 */
-fn linear_log_recur(n: f32) -> i32 {
-    if n <= 1.0 {
+fn linear_log_recur(n: i32) -> i32 {
+    if n <= 1 {
         return 1;
     }
-    let mut count = linear_log_recur(n / 2.0) + linear_log_recur(n / 2.0);
+    let mut count = linear_log_recur(n / 2) + linear_log_recur(n / 2);
     for _ in 0..n as i32 {
         count += 1;
     }
@@ -157,12 +157,12 @@ fn main() {
     count = exp_recur(n);
     println!("指数阶（递归实现）的操作数量 = {}", count);
 
-    count = logarithmic(n as f32);
+    count = logarithmic(n);
     println!("对数阶（循环实现）的操作数量 = {}", count);
-    count = log_recur(n as f32);
+    count = log_recur(n);
     println!("对数阶（递归实现）的操作数量 = {}", count);
 
-    count = linear_log_recur(n as f32);
+    count = linear_log_recur(n);
     println!("线性对数阶（递归实现）的操作数量 = {}", count);
 
     count = factorial_recur(n);