Add Java codes, and license.

This commit is contained in:
krahets
2022-11-08 02:58:42 +08:00
parent e604470259
commit 8f8f6319af
15 changed files with 952 additions and 17 deletions

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package chapter_computational_complexity.time_complexity;
public class time_complexity_types {
/* 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) {
// System.out.println(num);
count++;
}
return count;
}
/* 平方阶 */
static int quadratic(int n) {
int count = 0;
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 void bubbleSort(int[] nums) {
int n = nums.length;
for (int i = 0; i < n - 1; i++) {
for (int j = 0; j < n - 1 - i; j++) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 和 nums[j + 1]
int tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
}
}
}
}
/* 指数阶(循环实现) */
static int exponential(int n) {
int count = 0, base = 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 < 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);
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);
}
}

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package chapter_computational_complexity.time_complexity;
import java.util.*;
public class worst_best_time_complexity {
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
static int[] randomNumbers(int n) {
Integer[] nums = new Integer[n];
// 生成数组 nums = { 1, 2, 3, ..., n }
for (int i = 0; i < n; i++) {
nums[i] = i + 1;
}
// 随机打乱数组元素
Collections.shuffle(Arrays.asList(nums));
// Integer[] -> int[]
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nums[i];
}
return res;
}
/* 查找数组 nums 中数字 1 所在索引 */
static int findOne(int[] nums) {
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 1)
return i;
}
return -1;
}
/* Driver Code */
public static void main(String[] args) {
for (int i = 0; i < 10; i++) {
int n = 100;
int[] nums = randomNumbers(n);
int index = findOne(nums);
System.out.println("\n数组 [ 1, 2, ..., n ] 被打乱后 = " + Arrays.toString(nums));
System.out.println("数字 1 的索引为 " + index);
}
}
}