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2626de8d0b
introduction, computational complexity.
194 lines
5.6 KiB
C#
194 lines
5.6 KiB
C#
/**
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* File: time_complexity.cs
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* Created Time: 2022-12-23
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* Author: haptear (haptear@hotmail.com)
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*/
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namespace hello_algo.chapter_computational_complexity;
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public class time_complexity {
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void algorithm(int n) {
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int a = 1; // +0(技巧 1)
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a = a + n; // +0(技巧 1)
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// +n(技巧 2)
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for (int i = 0; i < 5 * n + 1; i++) {
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Console.WriteLine(0);
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}
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// +n*n(技巧 3)
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for (int i = 0; i < 2 * n; i++) {
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for (int j = 0; j < n + 1; j++) {
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Console.WriteLine(0);
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}
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}
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}
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// 算法 A 时间复杂度:常数阶
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void algorithm_A(int n) {
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Console.WriteLine(0);
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}
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// 算法 B 时间复杂度:线性阶
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void algorithm_B(int n) {
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for (int i = 0; i < n; i++) {
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Console.WriteLine(0);
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}
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}
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// 算法 C 时间复杂度:常数阶
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void algorithm_C(int n) {
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for (int i = 0; i < 1000000; i++) {
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Console.WriteLine(0);
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}
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}
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/* 常数阶 */
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static int constant(int n) {
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int count = 0;
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int size = 100000;
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for (int i = 0; i < size; i++)
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count++;
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return count;
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}
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/* 线性阶 */
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static int linear(int n) {
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int count = 0;
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for (int i = 0; i < n; i++)
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count++;
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return count;
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}
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/* 线性阶(遍历数组) */
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static int arrayTraversal(int[] nums) {
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int count = 0;
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// 循环次数与数组长度成正比
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foreach (int num in nums) {
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count++;
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}
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return count;
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}
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/* 平方阶 */
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static int quadratic(int n) {
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int count = 0;
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// 循环次数与数组长度成平方关系
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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count++;
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}
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}
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return count;
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}
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/* 平方阶(冒泡排序) */
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static int bubbleSort(int[] nums) {
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int count = 0; // 计数器
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// 外循环:未排序区间为 [0, i]
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for (int i = nums.Length - 1; i > 0; i--) {
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// 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
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for (int j = 0; j < i; j++) {
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if (nums[j] > nums[j + 1]) {
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// 交换 nums[j] 与 nums[j + 1]
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(nums[j + 1], nums[j]) = (nums[j], nums[j + 1]);
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count += 3; // 元素交换包含 3 个单元操作
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}
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}
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}
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return count;
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}
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/* 指数阶(循环实现) */
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static int exponential(int n) {
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int count = 0, bas = 1;
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// 细胞每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < bas; j++) {
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count++;
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}
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bas *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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/* 指数阶(递归实现) */
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static int expRecur(int n) {
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if (n == 1) return 1;
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return expRecur(n - 1) + expRecur(n - 1) + 1;
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}
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/* 对数阶(循环实现) */
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static int logarithmic(float n) {
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int count = 0;
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while (n > 1) {
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n = n / 2;
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count++;
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}
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return count;
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}
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/* 对数阶(递归实现) */
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static int logRecur(float n) {
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if (n <= 1) return 0;
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return logRecur(n / 2) + 1;
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}
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/* 线性对数阶 */
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static int linearLogRecur(float n) {
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if (n <= 1) return 1;
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int count = linearLogRecur(n / 2) +
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linearLogRecur(n / 2);
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for (int i = 0; i < n; i++) {
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count++;
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}
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return count;
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}
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/* 阶乘阶(递归实现) */
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static int factorialRecur(int n) {
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if (n == 0) return 1;
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int count = 0;
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// 从 1 个分裂出 n 个
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for (int i = 0; i < n; i++) {
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count += factorialRecur(n - 1);
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}
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return count;
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}
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[Test]
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public void Test() {
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// 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
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int n = 8;
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Console.WriteLine("输入数据大小 n = " + n);
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int count = constant(n);
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Console.WriteLine("常数阶的操作数量 = " + count);
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count = linear(n);
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Console.WriteLine("线性阶的操作数量 = " + count);
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count = arrayTraversal(new int[n]);
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Console.WriteLine("线性阶(遍历数组)的操作数量 = " + count);
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count = quadratic(n);
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Console.WriteLine("平方阶的操作数量 = " + count);
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int[] nums = new int[n];
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for (int i = 0; i < n; i++)
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nums[i] = n - i; // [n,n-1,...,2,1]
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count = bubbleSort(nums);
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Console.WriteLine("平方阶(冒泡排序)的操作数量 = " + count);
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count = exponential(n);
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Console.WriteLine("指数阶(循环实现)的操作数量 = " + count);
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count = expRecur(n);
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Console.WriteLine("指数阶(递归实现)的操作数量 = " + count);
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count = logarithmic((float)n);
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Console.WriteLine("对数阶(循环实现)的操作数量 = " + count);
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count = logRecur((float)n);
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Console.WriteLine("对数阶(递归实现)的操作数量 = " + count);
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count = linearLogRecur((float)n);
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Console.WriteLine("线性对数阶(递归实现)的操作数量 = " + count);
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count = factorialRecur(n);
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Console.WriteLine("阶乘阶(递归实现)的操作数量 = " + count);
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}
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}
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