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	3ea91bda99
	
	
	
		
			
			* Use int instead of float for the example code of log time complexity * Bug fixes * Bug fixes
		
			
				
	
	
		
			166 lines
		
	
	
		
			3.7 KiB
		
	
	
	
		
			Dart
		
	
	
	
	
	
			
		
		
	
	
			166 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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| 
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| // ignore_for_file: unused_local_variable
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| 
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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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| 
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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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| 
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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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| 
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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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|   // 循环次数与数据大小 n 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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| 
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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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| 
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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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| 
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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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| 
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| /* 对数阶(循环实现) */
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| int logarithmic(int 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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| 
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| /* 对数阶(递归实现) */
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| int logRecur(int 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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| 
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| /* 线性对数阶 */
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| int linearLogRecur(int 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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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