hello-algo/codes/java/chapter_computational_complexity/time_complexity.java
krahets 932d14644d Polish the content
Polish the chapter preface, introduction and complexity anlysis
2023-08-08 23:16:33 +08:00

168 lines
4.9 KiB
Java

/**
* File: time_complexity.java
* Created Time: 2022-11-25
* Author: Krahets (krahets@163.com)
*/
package chapter_computational_complexity;
public class time_complexity {
/* 常数阶 */
static int constant(int n) {
int count = 0;
int size = 100000;
for (int i = 0; i < size; i++)
count++;
return count;
}
/* 线性阶 */
static int linear(int n) {
int count = 0;
for (int i = 0; i < n; i++)
count++;
return count;
}
/* 线性阶(遍历数组) */
static int arrayTraversal(int[] nums) {
int count = 0;
// 循环次数与数组长度成正比
for (int num : nums) {
count++;
}
return count;
}
/* 平方阶 */
static 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) */
static int bubbleSort(int[] nums) {
int count = 0; // 计数器
// 外循环:未排序区间为 [0, i]
for (int i = nums.length - 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) */
static int exponential(int n) {
int count = 0, base = 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 < base; j++) {
count++;
}
base *= 2;
}
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
return count;
}
/* 指数阶(递归实现) */
static int expRecur(int n) {
if (n == 1)
return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
/* 对数阶(循环实现) */
static int logarithmic(float n) {
int count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
/* 对数阶(递归实现) */
static int logRecur(float n) {
if (n <= 1)
return 0;
return logRecur(n / 2) + 1;
}
/* 线性对数阶 */
static 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) */
static int factorialRecur(int n) {
if (n == 0)
return 1;
int count = 0;
// 从 1 个分裂出 n 个
for (int i = 0; i < n; i++) {
count += factorialRecur(n - 1);
}
return count;
}
/* Driver Code */
public static void main(String[] args) {
// 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;
System.out.println("输入数据大小 n = " + n);
int count = constant(n);
System.out.println("常数阶的计算操作数量 = " + count);
count = linear(n);
System.out.println("线性阶的计算操作数量 = " + count);
count = arrayTraversal(new int[n]);
System.out.println("线性阶(遍历数组)的计算操作数量 = " + count);
count = quadratic(n);
System.out.println("平方阶的计算操作数量 = " + count);
int[] nums = new int[n];
for (int i = 0; i < n; i++)
nums[i] = n - i; // [n,n-1,...,2,1]
count = bubbleSort(nums);
System.out.println("平方阶(冒泡排序)的计算操作数量 = " + count);
count = exponential(n);
System.out.println("指数阶(循环实现)的计算操作数量 = " + count);
count = expRecur(n);
System.out.println("指数阶(递归实现)的计算操作数量 = " + count);
count = logarithmic((float) n);
System.out.println("对数阶(循环实现)的计算操作数量 = " + count);
count = logRecur((float) n);
System.out.println("对数阶(递归实现)的计算操作数量 = " + count);
count = linearLogRecur((float) n);
System.out.println("线性对数阶(递归实现)的计算操作数量 = " + count);
count = factorialRecur(n);
System.out.println("阶乘阶(递归实现)的计算操作数量 = " + count);
}
}