[Rust] make rust part more idomatic and fix panic of backtrack template (#1370)

* Drop unused variable

* Idiomatic rust

* Fix panic template
This commit is contained in:
rongyi 2024-05-24 16:21:17 +08:00 committed by GitHub
parent aa818945f0
commit 63bcdb798a
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GPG key ID: B5690EEEBB952194
5 changed files with 27 additions and 28 deletions

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@ -74,7 +74,6 @@ fn find(nums: &[i32], target: i32) -> Option<usize> {
fn main() {
/* 初始化数组 */
let arr: [i32; 5] = [0; 5];
let slice: &[i32] = &[0; 5];
print!("数组 arr = ");
print_util::print_array(&arr);
// 在 Rust 中,指定长度时([i32; 5])为数组,不指定长度时(&[i32])为切片

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@ -10,7 +10,7 @@ use std::{cell::RefCell, rc::Rc};
use tree_node::{vec_to_tree, TreeNode};
/* 前序遍历:例题一 */
fn pre_order(res: &mut Vec<Rc<RefCell<TreeNode>>>, root: Option<Rc<RefCell<TreeNode>>>) {
fn pre_order(res: &mut Vec<Rc<RefCell<TreeNode>>>, root: Option<&Rc<RefCell<TreeNode>>>) {
if root.is_none() {
return;
}
@ -19,8 +19,8 @@ fn pre_order(res: &mut Vec<Rc<RefCell<TreeNode>>>, root: Option<Rc<RefCell<TreeN
// 记录解
res.push(node.clone());
}
pre_order(res, node.borrow().left.clone());
pre_order(res, node.borrow().right.clone());
pre_order(res, node.borrow().left.as_ref());
pre_order(res, node.borrow().right.as_ref());
}
}
@ -32,7 +32,7 @@ pub fn main() {
// 前序遍历
let mut res = Vec::new();
pre_order(&mut res, root);
pre_order(&mut res, root.as_ref());
println!("\n输出所有值为 7 的节点");
let mut vals = Vec::new();

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@ -13,7 +13,7 @@ use tree_node::{vec_to_tree, TreeNode};
fn pre_order(
res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
path: &mut Vec<Rc<RefCell<TreeNode>>>,
root: Option<Rc<RefCell<TreeNode>>>,
root: Option<&Rc<RefCell<TreeNode>>>,
) {
if root.is_none() {
return;
@ -25,10 +25,10 @@ fn pre_order(
// 记录解
res.push(path.clone());
}
pre_order(res, path, node.borrow().left.clone());
pre_order(res, path, node.borrow().right.clone());
pre_order(res, path, node.borrow().left.as_ref());
pre_order(res, path, node.borrow().right.as_ref());
// 回退
path.remove(path.len() - 1);
path.pop();
}
}
@ -41,7 +41,7 @@ pub fn main() {
// 前序遍历
let mut path = Vec::new();
let mut res = Vec::new();
pre_order(&mut res, &mut path, root);
pre_order(&mut res, &mut path, root.as_ref());
println!("\n输出所有根节点到节点 7 的路径");
for path in res {

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@ -13,7 +13,7 @@ use tree_node::{vec_to_tree, TreeNode};
fn pre_order(
res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
path: &mut Vec<Rc<RefCell<TreeNode>>>,
root: Option<Rc<RefCell<TreeNode>>>,
root: Option<&Rc<RefCell<TreeNode>>>,
) {
// 剪枝
if root.is_none() || root.as_ref().unwrap().borrow().val == 3 {
@ -26,10 +26,10 @@ fn pre_order(
// 记录解
res.push(path.clone());
}
pre_order(res, path, node.borrow().left.clone());
pre_order(res, path, node.borrow().right.clone());
pre_order(res, path, node.borrow().left.as_ref());
pre_order(res, path, node.borrow().right.as_ref());
// 回退
path.remove(path.len() - 1);
path.pop();
}
}
@ -42,7 +42,7 @@ pub fn main() {
// 前序遍历
let mut path = Vec::new();
let mut res = Vec::new();
pre_order(&mut res, &mut path, root);
pre_order(&mut res, &mut path, root.as_ref());
println!("\n输出所有根节点到节点 7 的路径,路径中不包含值为 3 的节点");
for path in res {

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@ -11,7 +11,7 @@ use tree_node::{vec_to_tree, TreeNode};
/* 判断当前状态是否为解 */
fn is_solution(state: &mut Vec<Rc<RefCell<TreeNode>>>) -> bool {
return !state.is_empty() && state.get(state.len() - 1).unwrap().borrow().val == 7;
return !state.is_empty() && state.last().unwrap().borrow().val == 7;
}
/* 记录解 */
@ -23,8 +23,8 @@ fn record_solution(
}
/* 判断在当前状态下,该选择是否合法 */
fn is_valid(_: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Rc<RefCell<TreeNode>>) -> bool {
return choice.borrow().val != 3;
fn is_valid(_: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Option<&Rc<RefCell<TreeNode>>>) -> bool {
return choice.is_some() && choice.unwrap().borrow().val != 3;
}
/* 更新状态 */
@ -34,13 +34,13 @@ fn make_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Rc<RefCell<TreeNo
/* 恢复状态 */
fn undo_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, _: Rc<RefCell<TreeNode>>) {
state.remove(state.len() - 1);
state.pop();
}
/* 回溯算法:例题三 */
fn backtrack(
state: &mut Vec<Rc<RefCell<TreeNode>>>,
choices: &mut Vec<Rc<RefCell<TreeNode>>>,
choices: &Vec<Option<&Rc<RefCell<TreeNode>>>>,
res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
) {
// 检查是否为解
@ -49,22 +49,22 @@ fn backtrack(
record_solution(state, res);
}
// 遍历所有选择
for choice in choices {
for &choice in choices.iter() {
// 剪枝:检查选择是否合法
if is_valid(state, choice.clone()) {
if is_valid(state, choice) {
// 尝试:做出选择,更新状态
make_choice(state, choice.clone());
make_choice(state, choice.unwrap().clone());
// 进行下一轮选择
backtrack(
state,
&mut vec![
choice.borrow().left.clone().unwrap(),
choice.borrow().right.clone().unwrap(),
&vec![
choice.unwrap().borrow().left.as_ref(),
choice.unwrap().borrow().right.as_ref(),
],
res,
);
// 回退:撤销选择,恢复到之前的状态
undo_choice(state, choice.clone());
undo_choice(state, choice.unwrap().clone());
}
}
}
@ -77,7 +77,7 @@ pub fn main() {
// 回溯算法
let mut res = Vec::new();
backtrack(&mut Vec::new(), &mut vec![root.unwrap()], &mut res);
backtrack(&mut Vec::new(), &mut vec![root.as_ref()], &mut res);
println!("\n输出所有根节点到节点 7 的路径,要求路径中不包含值为 3 的节点");
for path in res {