mirror of
				https://github.com/krahets/hello-algo.git
				synced 2025-11-01 03:24:24 +08:00 
			
		
		
		
	 3ea91bda99
			
		
	
	3ea91bda99
	
	
	
		
			
			* Use int instead of float for the example code of log time complexity * Bug fixes * Bug fixes
		
			
				
	
	
		
			180 lines
		
	
	
		
			4.3 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
			
		
		
	
	
			180 lines
		
	
	
		
			4.3 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
| /**
 | ||
|  * File: time_complexity.c
 | ||
|  * Created Time: 2023-01-03
 | ||
|  * Author: codingonion (coderonion@gmail.com)
 | ||
|  */
 | ||
| 
 | ||
| #include "../utils/common.h"
 | ||
| 
 | ||
| /* 常数阶 */
 | ||
| int constant(int n) {
 | ||
|     int count = 0;
 | ||
|     int size = 100000;
 | ||
|     int i = 0;
 | ||
|     for (int i = 0; i < size; i++) {
 | ||
|         count++;
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 线性阶 */
 | ||
| int linear(int n) {
 | ||
|     int count = 0;
 | ||
|     for (int i = 0; i < n; i++) {
 | ||
|         count++;
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 线性阶(遍历数组) */
 | ||
| int arrayTraversal(int *nums, int n) {
 | ||
|     int count = 0;
 | ||
|     // 循环次数与数组长度成正比
 | ||
|     for (int i = 0; i < n; i++) {
 | ||
|         count++;
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 平方阶 */
 | ||
| int quadratic(int n) {
 | ||
|     int count = 0;
 | ||
|     // 循环次数与数据大小 n 成平方关系
 | ||
|     for (int i = 0; i < n; i++) {
 | ||
|         for (int j = 0; j < n; j++) {
 | ||
|             count++;
 | ||
|         }
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 平方阶(冒泡排序) */
 | ||
| int bubbleSort(int *nums, int n) {
 | ||
|     int count = 0; // 计数器
 | ||
|     // 外循环:未排序区间为 [0, i]
 | ||
|     for (int i = n - 1; i > 0; i--) {
 | ||
|         // 内循环:将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
 | ||
|         for (int j = 0; j < i; j++) {
 | ||
|             if (nums[j] > nums[j + 1]) {
 | ||
|                 // 交换 nums[j] 与 nums[j + 1]
 | ||
|                 int tmp = nums[j];
 | ||
|                 nums[j] = nums[j + 1];
 | ||
|                 nums[j + 1] = tmp;
 | ||
|                 count += 3; // 元素交换包含 3 个单元操作
 | ||
|             }
 | ||
|         }
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 指数阶(循环实现) */
 | ||
| int exponential(int n) {
 | ||
|     int count = 0;
 | ||
|     int bas = 1;
 | ||
|     // 细胞每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
 | ||
|     for (int i = 0; i < n; i++) {
 | ||
|         for (int j = 0; j < bas; j++) {
 | ||
|             count++;
 | ||
|         }
 | ||
|         bas *= 2;
 | ||
|     }
 | ||
|     // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 指数阶(递归实现) */
 | ||
| int expRecur(int n) {
 | ||
|     if (n == 1)
 | ||
|         return 1;
 | ||
|     return expRecur(n - 1) + expRecur(n - 1) + 1;
 | ||
| }
 | ||
| 
 | ||
| /* 对数阶(循环实现) */
 | ||
| int logarithmic(int n) {
 | ||
|     int count = 0;
 | ||
|     while (n > 1) {
 | ||
|         n = n / 2;
 | ||
|         count++;
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 对数阶(递归实现) */
 | ||
| int logRecur(int n) {
 | ||
|     if (n <= 1)
 | ||
|         return 0;
 | ||
|     return logRecur(n / 2) + 1;
 | ||
| }
 | ||
| 
 | ||
| /* 线性对数阶 */
 | ||
| int linearLogRecur(int n) {
 | ||
|     if (n <= 1)
 | ||
|         return 1;
 | ||
|     int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
 | ||
|     for (int i = 0; i < n; i++) {
 | ||
|         count++;
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* 阶乘阶(递归实现) */
 | ||
| int factorialRecur(int n) {
 | ||
|     if (n == 0)
 | ||
|         return 1;
 | ||
|     int count = 0;
 | ||
|     for (int i = 0; i < n; i++) {
 | ||
|         count += factorialRecur(n - 1);
 | ||
|     }
 | ||
|     return count;
 | ||
| }
 | ||
| 
 | ||
| /* Driver Code */
 | ||
| int main(int argc, char *argv[]) {
 | ||
|     // 可以修改 n 运行,体会一下各种复杂度的操作数量变化趋势
 | ||
|     int n = 8;
 | ||
|     printf("输入数据大小 n = %d\n", n);
 | ||
| 
 | ||
|     int count = constant(n);
 | ||
|     printf("常数阶的操作数量 = %d\n", count);
 | ||
| 
 | ||
|     count = linear(n);
 | ||
|     printf("线性阶的操作数量 = %d\n", count);
 | ||
|     // 分配堆区内存(创建一维可变长数组:数组中元素数量为 n ,元素类型为 int )
 | ||
|     int *nums = (int *)malloc(n * sizeof(int));
 | ||
|     count = arrayTraversal(nums, n);
 | ||
|     printf("线性阶(遍历数组)的操作数量 = %d\n", count);
 | ||
| 
 | ||
|     count = quadratic(n);
 | ||
|     printf("平方阶的操作数量 = %d\n", count);
 | ||
|     for (int i = 0; i < n; i++) {
 | ||
|         nums[i] = n - i; // [n,n-1,...,2,1]
 | ||
|     }
 | ||
|     count = bubbleSort(nums, n);
 | ||
|     printf("平方阶(冒泡排序)的操作数量 = %d\n", count);
 | ||
| 
 | ||
|     count = exponential(n);
 | ||
|     printf("指数阶(循环实现)的操作数量 = %d\n", count);
 | ||
|     count = expRecur(n);
 | ||
|     printf("指数阶(递归实现)的操作数量 = %d\n", count);
 | ||
| 
 | ||
|     count = logarithmic(n);
 | ||
|     printf("对数阶(循环实现)的操作数量 = %d\n", count);
 | ||
|     count = logRecur(n);
 | ||
|     printf("对数阶(递归实现)的操作数量 = %d\n", count);
 | ||
| 
 | ||
|     count = linearLogRecur(n);
 | ||
|     printf("线性对数阶(递归实现)的操作数量 = %d\n", count);
 | ||
| 
 | ||
|     count = factorialRecur(n);
 | ||
|     printf("阶乘阶(递归实现)的操作数量 = %d\n", count);
 | ||
| 
 | ||
|     // 释放堆区内存
 | ||
|     if (nums != NULL) {
 | ||
|         free(nums);
 | ||
|         nums = NULL;
 | ||
|     }
 | ||
|     getchar();
 | ||
| 
 | ||
|     return 0;
 | ||
| }
 |