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✨ feat(code): add rust time_complexity.rs example
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163
codes/rust/chapter_computational_complexity/time_complexity.rs
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163
codes/rust/chapter_computational_complexity/time_complexity.rs
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#![allow(unused_variables)]
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/* 常数阶 */
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fn constant(n: i32) -> i32 {
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let mut count = 0;
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let size = 100000;
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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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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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fn quadratic(n: i32) -> i32 {
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let mut 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 _ 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:待排序元素数量为 n-1, n-2, ..., 1
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for i in (1..nums.len()).rev() {
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// 内循环:冒泡操作
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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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// cell 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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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: f64) -> i32 {
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if n <= 1.0 {
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return 1;
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}
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let mut count = linear_log_recur(n / 2.0) + linear_log_recur(n / 2.0);
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for _ in 0 ..n as i32 {
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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!("常数阶的计算操作数量 = {}", count);
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count = linear(n);
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println!("线性阶的计算操作数量 = {}", 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)的计算操作数量 = {}", count);
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count = quadratic(n);
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println!("平方阶的计算操作数量 = {}", 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)的计算操作数量 = {}", count);
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count = exponential(n);
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println!("指数阶(循环实现)的计算操作数量 = {}", count);
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count = exp_recur(n);
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println!("指数阶(递归实现)的计算操作数量 = {}", count);
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count = logarithmic(n);
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println!("对数阶(循环实现)的计算操作数量 = {}", count);
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count = log_recur(n);
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println!("对数阶(递归实现)的计算操作数量 = {}", count);
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count = linear_log_recur(n.into());
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println!("线性对数阶(递归实现)的计算操作数量 = {}", count);
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count = factorial_recur(n);
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println!("阶乘阶(递归实现)的计算操作数量 = {}", count);
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}
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