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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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/* 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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/* 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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/* 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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/* 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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/* 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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    // 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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/* 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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/* 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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/* 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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/* 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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/* 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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/* 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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/* 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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    let mut count = constant(n);
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    println!("常数阶的操作数量 = {}", count);
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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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    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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    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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    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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    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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    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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