hello-algo/codes/c/chapter_computational_complexity/time_complexity.c

176 lines
4.3 KiB
C
Raw Normal View History

/*
* File: time_complexity.c
* Created Time: 2023-01-03
* Author: sjinzh (sjinzh@gmail.com)
*/
#include "../include/include.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;
// 循环次数与数组长度成平方关系
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:待排序元素数量为 n-1, n-2, ..., 1
for (int i = n - 1; i > 0; i--) {
// 内循环:冒泡操作
for (int j = 0; j < i; j++) {
// 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;
// cell 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(float n) {
int count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
/* 对数阶(递归实现) */
int logRecur(float n) {
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
/* 线性对数阶 */
int linearLogRecur(float 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("常数阶的计算操作数量 = %d\n", count);
count = linear(n);
printf("线性阶的计算操作数量 = %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)的计算操作数量 = %d\n", count);
count = quadratic(n);
printf("平方阶的计算操作数量 = %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)的计算操作数量 = %d\n", count);
count = exponential(n);
printf("指数阶(循环实现)的计算操作数量 = %d\n", count);
count = expRecur(n);
printf("指数阶(递归实现)的计算操作数量 = %d\n", count);
count = logarithmic(n);
printf("对数阶(循环实现)的计算操作数量 = %d\n", count);
count = logRecur(n);
printf("对数阶(递归实现)的计算操作数量 = %d\n", count);
count = linearLogRecur(n);
printf("线性对数阶(递归实现)的计算操作数量 = %d\n", count);
count = factorialRecur(n);
printf("阶乘阶(递归实现)的计算操作数量 = %d\n", count);
// 释放堆区内存
if (nums != NULL) {
free(nums);
nums = NULL;
}
getchar();
return 0;
}