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			171 lines
		
	
	
		
			4.0 KiB
		
	
	
	
		
			Rust
		
	
	
	
	
	
			
		
		
	
	
			171 lines
		
	
	
		
			4.0 KiB
		
	
	
	
		
			Rust
		
	
	
	
	
	
| /*
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|  * File: time_complexity.rs
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|  * Created Time: 2023-01-10
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|  * Author: xBLACICEx (xBLACKICEx@outlook.com), codingonion (coderonion@gmail.com)
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|  */
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| 
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| /* 常数阶 */
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| fn constant(n: i32) -> i32 {
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|     _ = n;
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|     let mut count = 0;
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|     let size = 100_000;
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|     for _ in 0..size {
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|         count += 1;
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|     }
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|     count
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| }
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| 
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| /* 线性阶 */
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| fn linear(n: i32) -> i32 {
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|     let mut count = 0;
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|     for _ in 0..n {
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|         count += 1;
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|     }
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|     count
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| }
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| 
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| /* 线性阶(遍历数组) */
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| fn array_traversal(nums: &[i32]) -> i32 {
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|     let mut count = 0;
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|     // 循环次数与数组长度成正比
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|     for _ in nums {
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|         count += 1;
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|     }
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|     count
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| }
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| 
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| /* 平方阶 */
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| fn quadratic(n: i32) -> i32 {
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|     let mut 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 _ in 0..n {
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|         for _ in 0..n {
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|             count += 1;
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|         }
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|     }
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|     count
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| }
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| 
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| /* 平方阶(冒泡排序) */
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| fn bubble_sort(nums: &mut [i32]) -> i32 {
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|     let mut count = 0; // 计数器
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| 
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|     // 外循环:未排序区间为 [0, i]
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|     for i in (1..nums.len()).rev() {
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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 j in 0..i {
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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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|                 let 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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|     count
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| }
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| 
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| /* 指数阶(循环实现) */
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| fn exponential(n: i32) -> i32 {
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|     let mut count = 0;
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|     let mut 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 _ in 0..n {
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|         for _ in 0..base {
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|             count += 1
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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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|     count
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| }
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| 
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| /* 指数阶(递归实现) */
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| fn exp_recur(n: i32) -> i32 {
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|     if n == 1 {
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|         return 1;
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|     }
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|     exp_recur(n - 1) + exp_recur(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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| fn logarithmic(mut n: i32) -> i32 {
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|     let mut count = 0;
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|     while n > 1 {
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|         n = n / 2;
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|         count += 1;
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|     }
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|     count
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| }
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| 
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| /* 对数阶(递归实现) */
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| fn log_recur(n: i32) -> i32 {
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|     if n <= 1 {
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|         return 0;
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|     }
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|     log_recur(n / 2) + 1
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| }
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| 
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| /* 线性对数阶 */
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| fn linear_log_recur(n: i32) -> i32 {
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|     if n <= 1 {
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|         return 1;
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|     }
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|     let mut count = linear_log_recur(n / 2) + linear_log_recur(n / 2);
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|     for _ in 0..n {
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|         count += 1;
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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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| fn factorial_recur(n: i32) -> i32 {
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|     if n == 0 {
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|         return 1;
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|     }
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|     let mut count = 0;
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|     // 从 1 个分裂出 n 个
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|     for _ in 0..n {
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|         count += factorial_recur(n - 1);
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|     }
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|     count
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| }
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| 
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| /* Driver Code */
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| fn 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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|     let n: i32 = 8;
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|     println!("输入数据大小 n = {}", n);
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| 
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|     let mut count = constant(n);
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|     println!("常数阶的操作数量 = {}", count);
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| 
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|     count = linear(n);
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|     println!("线性阶的操作数量 = {}", count);
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|     count = array_traversal(&vec![0; n as usize]);
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|     println!("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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|     println!("平方阶的操作数量 = {}", count);
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|     let mut nums = (1..=n).rev().collect::<Vec<_>>(); // [n,n-1,...,2,1]
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|     count = bubble_sort(&mut nums);
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|     println!("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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|     println!("指数阶(循环实现)的操作数量 = {}", count);
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|     count = exp_recur(n);
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|     println!("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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|     println!("对数阶(循环实现)的操作数量 = {}", count);
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|     count = log_recur(n);
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|     println!("对数阶(递归实现)的操作数量 = {}", count);
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| 
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|     count = linear_log_recur(n);
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|     println!("线性对数阶(递归实现)的操作数量 = {}", count);
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| 
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|     count = factorial_recur(n);
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|     println!("阶乘阶(递归实现)的操作数量 = {}", count);
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| }
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