Merge pull request #251 from xBLACKICEx/rust-computational_complexity

Rust computational complexity
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Yudong Jin 2023-01-19 02:24:52 +08:00 committed by GitHub
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7 changed files with 200 additions and 1 deletions

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codes/rust/.gitignore vendored Normal file
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**/target
**/worst_best_time_complexity/
**/time_complexity/

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name = "chapter_computational_complexity"
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edition = "2021"
[[bin]]
name = "leetcode_two_sum"
path = "leetcode_two_sum.rs"
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name = "time_complexity"
path = "time_complexity.rs"
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/**
* File: leetcode_two_sum.rs
* Created Time: 2023-01-14
* Author: xBLACICEx (xBLACKICEx@outlook.com)
*/
use std::collections::HashMap;
struct SolutionBruteForce;
struct SolutionHashMap;
impl SolutionBruteForce {
pub fn two_sum(nums: &Vec<i32>, target: i32) -> Vec<i32> {
for i in 0..nums.len() - 1 {
for j in i + 1..nums.len() {
if nums[i] + nums[j] == target {
return vec![i as i32, j as i32];
}
}
}
vec![]
}
}
impl SolutionHashMap {
pub fn two_sum(nums: &Vec<i32>, target: i32) -> Vec<i32> {
let mut hm = HashMap::new();
for (i, n) in nums.iter().enumerate() {
match hm.get(&(target - n)) {
Some(v) => return vec![*v as i32, i as i32],
None => hm.insert(n, i)
};
}
vec![]
}
}
// Driver Code
fn main() {
// ======= Test Case =======
let nums = vec![2,7,11,15];
let target = 9;
// 方法一
let res = SolutionBruteForce::two_sum(&nums, target);
println!("方法一 res = {:?}", res);
// 方法二
let res = SolutionHashMap::two_sum(&nums, target);
println!("方法二 res = {:?}", res);
}

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#![allow(unused_variables)] /**
* File: time_complexity.rs
* Created Time: 2023-01-10
* Author: xBLACICEx (xBLACKICEx@outlook.com)
*/
#[allow(unused_variables)]
/* 常数阶 */ /* 常数阶 */
fn constant(n: i32) -> i32 { fn constant(n: i32) -> i32 {
@ -10,6 +15,7 @@ fn constant(n: i32) -> i32 {
count count
} }
/* 线性阶 */
fn linear(n: i32) -> i32 { fn linear(n: i32) -> i32 {
let mut count = 0; let mut count = 0;
for _ in 0..n { for _ in 0..n {
@ -92,6 +98,7 @@ fn logarithmic(mut n: i32) -> i32 {
count count
} }
/* 对数阶(递归实现) */
fn log_recur(n: i32) -> i32 { fn log_recur(n: i32) -> i32 {
if n <= 1 { if n <= 1 {
return 0; return 0;

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/**
* File: time_complexity.rs
* Created Time: 2023-01-13
* Author: xBLACICEx (xBLACKICEx@outlook.com)
*/
use rand::seq::SliceRandom;
use rand::thread_rng;
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
fn random_numbers(n: i32) -> Vec<i32> {
// 生成数组 nums = { 1, 2, 3, ..., n }
let mut nums = (1..n + 1).collect::<Vec<i32>>();
// 随机打乱数组元素
nums.shuffle(&mut thread_rng());
nums
}
/* 查找数组 nums 中数字 1 所在索引 */
fn find_one(nums: &[i32]) -> Option<usize> {
for i in 0..nums.len() {
if nums[i] == 1 {
return Some(i);
}
}
None
}
/* Driver Code */
fn main() {
for _ in 0..10 {
let n = 100;
let nums = random_numbers(n);
let index = find_one(&nums);
println!("\n数组 [ 1, 2, ..., n ] 被打乱后 = {:?}", nums);
println!("数字 1 的索引为 {:?}", index);
}
}