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a8b2350705
* feat(dart): Add build check for Dart * feat(dart): Add dart analyze to check * fix(dart): remove dart analyze * feat(dart): Ignore unused variable and add dart analyze
165 lines
3.7 KiB
Dart
165 lines
3.7 KiB
Dart
/**
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* File: time_complexity.dart
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* Created Time: 2023-02-12
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* Author: Jefferson (JeffersonHuang77@gmail.com)
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*/
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// ignore_for_file: unused_local_variable
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/* 常数阶 */
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int constant(int n) {
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int count = 0;
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int size = 100000;
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for (var i = 0; i < size; i++) {
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count++;
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}
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return count;
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}
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/* 线性阶 */
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int linear(int n) {
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int count = 0;
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for (var 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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/* 线性阶(遍历数组) */
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int arrayTraversal(List<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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for (var 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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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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int bubbleSort(List<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 (var 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 (var 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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int tmp = nums[j];
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nums[j] = nums[j + 1];
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nums[j + 1] = tmp;
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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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int exponential(int n) {
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int count = 0, base = 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 (var i = 0; i < n; i++) {
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for (var j = 0; j < base; j++) {
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count++;
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}
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base *= 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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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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int logarithmic(num 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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int logRecur(num 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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int linearLogRecur(num n) {
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if (n <= 1) return 1;
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int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
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for (var 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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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 (var 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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/* Driver Code */
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void main() {
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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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print('输入数据大小 n = $n');
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int count = constant(n);
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print('常数阶的操作数量 = $count');
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count = linear(n);
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print('线性阶的操作数量 = $count');
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count = arrayTraversal(List.filled(n, 0));
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print('线性阶(遍历数组)的操作数量 = $count');
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count = quadratic(n);
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print('平方阶的操作数量 = $count');
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final nums = List.filled(n, 0);
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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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}
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count = bubbleSort(nums);
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print('平方阶(冒泡排序)的操作数量 = $count');
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count = exponential(n);
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print('指数阶(循环实现)的操作数量 = $count');
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count = expRecur(n);
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print('指数阶(递归实现)的操作数量 = $count');
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count = logarithmic(n);
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print('对数阶(循环实现)的操作数量 = $count');
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count = logRecur(n);
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print('对数阶(递归实现)的操作数量 = $count');
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count = linearLogRecur(n);
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print('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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print('阶乘阶(递归实现)的操作数量 = $count');
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}
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