cargo fmt rust code (#1131)

* cargo fmt code * Add empty line to seperate unrelated comments * Fix review * Update bubble_sort.rs * Update merge_sort.rs --------- Co-authored-by: Yudong Jin <krahets@163.com>
2026-07-12 07:26:07 +00:00 · 2024-03-16 02:13:41 +08:00
parent 54ceef3443
commit 7b1094318b
70 changed files with 1021 additions and 836 deletions
@@ -5,8 +5,15 @@
 */

 /* 回溯算法：n 皇后 */
-fn backtrack(row: usize, n: usize, state: &mut Vec<Vec<String>>, res: &mut Vec<Vec<Vec<String>>>,
-    cols: &mut [bool], diags1: &mut [bool], diags2: &mut [bool]) {
+fn backtrack(
+    row: usize,
+    n: usize,
+    state: &mut Vec<Vec<String>>,
+    res: &mut Vec<Vec<Vec<String>>>,
+    cols: &mut [bool],
+    diags1: &mut [bool],
+    diags2: &mut [bool],
+) {
    // 当放置完所有行时，记录解
    if row == n {
        let mut copy_state: Vec<Vec<String>> = Vec::new();
@@ -51,7 +58,15 @@ fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
    let mut diags2 = vec![false; 2 * n - 1]; // 记录次对角线上是否有皇后
    let mut res: Vec<Vec<Vec<String>>> = Vec::new();

-    backtrack(0, n, &mut state, &mut res, &mut cols, &mut diags1, &mut diags2);
+    backtrack(
+        0,
+        n,
+        &mut state,
+        &mut res,
+        &mut cols,
+        &mut diags1,
+        &mut diags2,
+    );

    res
 }
@@ -37,7 +37,7 @@ fn permutations_i(nums: &mut [i32]) -> Vec<Vec<i32>> {

 /* Driver Code */
 pub fn main() {
-    let mut nums = [ 1, 2, 3 ];
+    let mut nums = [1, 2, 3];

    let res = permutations_i(&mut nums);

@@ -41,7 +41,7 @@ fn permutations_ii(nums: &mut [i32]) -> Vec<Vec<i32>> {

 /* Driver Code */
 pub fn main() {
-    let mut nums = [ 1, 2, 2 ];
+    let mut nums = [1, 2, 2];

    let res = permutations_ii(&mut nums);

@@ -1,50 +1,54 @@
-/*
- * File: preorder_traversal_ii_compact.rs
- * Created Time: 2023-07-15
- * Author: codingonion (coderonion@gmail.com)
- */
-
-include!("../include/include.rs");
-
-use std::{cell::RefCell, rc::Rc};
-use tree_node::{vec_to_tree, TreeNode};
-
-/* 前序遍历：例题二 */
-fn pre_order(res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>, path: &mut Vec<Rc<RefCell<TreeNode>>>, root: Option<Rc<RefCell<TreeNode>>>) {
-    if root.is_none() {
-        return;
-    }
-    if let Some(node) = root {
-        // 尝试
-        path.push(node.clone());
-        if node.borrow().val == 7 {
-            // 记录解
-            res.push(path.clone());
-        }
-        pre_order(res, path, node.borrow().left.clone());
-        pre_order(res, path, node.borrow().right.clone());
-        // 回退
-        path.remove(path.len() -  1);
-    }
-}
-
-/* Driver Code */
-pub fn main() {
-    let root = vec_to_tree([1, 7, 3, 4, 5, 6, 7].map(|x| Some(x)).to_vec());
-    println!("初始化二叉树");
-    print_util::print_tree(root.as_ref().unwrap());
-
-    // 前序遍历
-    let mut path = Vec::new();
-    let mut res = Vec::new();
-    pre_order(&mut res, &mut path, root);
-
-    println!("\n输出所有根节点到节点 7 的路径");
-    for path in res {
-        let mut vals = Vec::new();
-        for node in path {
-            vals.push(node.borrow().val)
-        }
-        println!("{:?}", vals);
-    }
-}
+/*
+ * File: preorder_traversal_ii_compact.rs
+ * Created Time: 2023-07-15
+ * Author: codingonion (coderonion@gmail.com)
+ */
+
+include!("../include/include.rs");
+
+use std::{cell::RefCell, rc::Rc};
+use tree_node::{vec_to_tree, TreeNode};
+
+/* 前序遍历：例题二 */
+fn pre_order(
+    res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
+    path: &mut Vec<Rc<RefCell<TreeNode>>>,
+    root: Option<Rc<RefCell<TreeNode>>>,
+) {
+    if root.is_none() {
+        return;
+    }
+    if let Some(node) = root {
+        // 尝试
+        path.push(node.clone());
+        if node.borrow().val == 7 {
+            // 记录解
+            res.push(path.clone());
+        }
+        pre_order(res, path, node.borrow().left.clone());
+        pre_order(res, path, node.borrow().right.clone());
+        // 回退
+        path.remove(path.len() - 1);
+    }
+}
+
+/* Driver Code */
+pub fn main() {
+    let root = vec_to_tree([1, 7, 3, 4, 5, 6, 7].map(|x| Some(x)).to_vec());
+    println!("初始化二叉树");
+    print_util::print_tree(root.as_ref().unwrap());
+
+    // 前序遍历
+    let mut path = Vec::new();
+    let mut res = Vec::new();
+    pre_order(&mut res, &mut path, root);
+
+    println!("\n输出所有根节点到节点 7 的路径");
+    for path in res {
+        let mut vals = Vec::new();
+        for node in path {
+            vals.push(node.borrow().val)
+        }
+        println!("{:?}", vals);
+    }
+}
@@ -1,51 +1,55 @@
-/*
- * File: preorder_traversal_iii_compact.rs
- * Created Time: 2023-07-15
- * Author: codingonion (coderonion@gmail.com)
- */
-
-include!("../include/include.rs");
-
-use std::{cell::RefCell, rc::Rc};
-use tree_node::{vec_to_tree, TreeNode};
-
-/* 前序遍历：例题三 */
-fn pre_order(res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>, path: &mut Vec<Rc<RefCell<TreeNode>>>, root: Option<Rc<RefCell<TreeNode>>>) {
-    // 剪枝
-    if root.is_none() || root.as_ref().unwrap().borrow().val == 3 {
-        return;
-    }
-    if let Some(node) = root {
-        // 尝试
-        path.push(node.clone());
-        if node.borrow().val == 7 {
-            // 记录解
-            res.push(path.clone());
-        }
-        pre_order(res, path, node.borrow().left.clone());
-        pre_order(res, path, node.borrow().right.clone());
-        // 回退
-        path.remove(path.len() -  1);
-    }
-}
-
-/* Driver Code */
-pub fn main() {
-    let root = vec_to_tree([1, 7, 3, 4, 5, 6, 7].map(|x| Some(x)).to_vec());
-    println!("初始化二叉树");
-    print_util::print_tree(root.as_ref().unwrap());
-
-    // 前序遍历
-    let mut path = Vec::new();
-    let mut res = Vec::new();
-    pre_order(&mut res, &mut path, root);
-
-    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 {
-        let mut vals = Vec::new();
-        for node in path {
-            vals.push(node.borrow().val)
-        }
-        println!("{:?}", vals);
-    }
-}
+/*
+ * File: preorder_traversal_iii_compact.rs
+ * Created Time: 2023-07-15
+ * Author: codingonion (coderonion@gmail.com)
+ */
+
+include!("../include/include.rs");
+
+use std::{cell::RefCell, rc::Rc};
+use tree_node::{vec_to_tree, TreeNode};
+
+/* 前序遍历：例题三 */
+fn pre_order(
+    res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
+    path: &mut Vec<Rc<RefCell<TreeNode>>>,
+    root: Option<Rc<RefCell<TreeNode>>>,
+) {
+    // 剪枝
+    if root.is_none() || root.as_ref().unwrap().borrow().val == 3 {
+        return;
+    }
+    if let Some(node) = root {
+        // 尝试
+        path.push(node.clone());
+        if node.borrow().val == 7 {
+            // 记录解
+            res.push(path.clone());
+        }
+        pre_order(res, path, node.borrow().left.clone());
+        pre_order(res, path, node.borrow().right.clone());
+        // 回退
+        path.remove(path.len() - 1);
+    }
+}
+
+/* Driver Code */
+pub fn main() {
+    let root = vec_to_tree([1, 7, 3, 4, 5, 6, 7].map(|x| Some(x)).to_vec());
+    println!("初始化二叉树");
+    print_util::print_tree(root.as_ref().unwrap());
+
+    // 前序遍历
+    let mut path = Vec::new();
+    let mut res = Vec::new();
+    pre_order(&mut res, &mut path, root);
+
+    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 {
+        let mut vals = Vec::new();
+        for node in path {
+            vals.push(node.borrow().val)
+        }
+        println!("{:?}", vals);
+    }
+}
@@ -1,76 +1,90 @@
-/*
- * File: preorder_traversal_iii_template.rs
- * Created Time: 2023-07-15
- * Author: codingonion (coderonion@gmail.com)
- */
-
-include!("../include/include.rs");
-
-use std::{cell::RefCell, rc::Rc};
-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;
-}
-
-/* 记录解 */
-fn record_solution(state: &mut Vec<Rc<RefCell<TreeNode>>>, res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>) {
-    res.push(state.clone());
-}
-
-/* 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;
-}
-
-/* 更新状态 */
-fn make_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Rc<RefCell<TreeNode>>) {
-    state.push(choice);
-}
-
-/* 恢复状态 */
-fn undo_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, _: Rc<RefCell<TreeNode>>) {
-    state.remove(state.len() - 1);
-}
-
-/* 回溯算法：例题三 */
-fn backtrack(state: &mut Vec<Rc<RefCell<TreeNode>>>, choices: &mut Vec<Rc<RefCell<TreeNode>>>, res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>) {
-    // 检查是否为解
-    if is_solution(state) {
-        // 记录解
-        record_solution(state, res);
-    }
-    // 遍历所有选择
-    for choice in choices {
-        // 剪枝：检查选择是否合法
-        if is_valid(state, choice.clone()) {
-            // 尝试：做出选择，更新状态
-            make_choice(state, choice.clone());
-            // 进行下一轮选择
-            backtrack(state, &mut vec![choice.borrow().left.clone().unwrap(), choice.borrow().right.clone().unwrap()], res);
-            // 回退：撤销选择，恢复到之前的状态
-            undo_choice(state, choice.clone());
-        }
-    }
-}
-
-/* Driver Code */
-pub fn main() {
-    let root = vec_to_tree([1, 7, 3, 4, 5, 6, 7].map(|x| Some(x)).to_vec());
-    println!("初始化二叉树");
-    print_util::print_tree(root.as_ref().unwrap());
-
-    // 回溯算法
-    let mut res = Vec::new();
-    backtrack(&mut Vec::new(), &mut vec![root.unwrap()], &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 {
-        let mut vals = Vec::new();
-        for node in path {
-            vals.push(node.borrow().val)
-        }
-        println!("{:?}", vals);
-    }
-}
+/*
+ * File: preorder_traversal_iii_template.rs
+ * Created Time: 2023-07-15
+ * Author: codingonion (coderonion@gmail.com)
+ */
+
+include!("../include/include.rs");
+
+use std::{cell::RefCell, rc::Rc};
+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;
+}
+
+/* 记录解 */
+fn record_solution(
+    state: &mut Vec<Rc<RefCell<TreeNode>>>,
+    res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
+) {
+    res.push(state.clone());
+}
+
+/* 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;
+}
+
+/* 更新状态 */
+fn make_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Rc<RefCell<TreeNode>>) {
+    state.push(choice);
+}
+
+/* 恢复状态 */
+fn undo_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, _: Rc<RefCell<TreeNode>>) {
+    state.remove(state.len() - 1);
+}
+
+/* 回溯算法：例题三 */
+fn backtrack(
+    state: &mut Vec<Rc<RefCell<TreeNode>>>,
+    choices: &mut Vec<Rc<RefCell<TreeNode>>>,
+    res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
+) {
+    // 检查是否为解
+    if is_solution(state) {
+        // 记录解
+        record_solution(state, res);
+    }
+    // 遍历所有选择
+    for choice in choices {
+        // 剪枝：检查选择是否合法
+        if is_valid(state, choice.clone()) {
+            // 尝试：做出选择，更新状态
+            make_choice(state, choice.clone());
+            // 进行下一轮选择
+            backtrack(
+                state,
+                &mut vec![
+                    choice.borrow().left.clone().unwrap(),
+                    choice.borrow().right.clone().unwrap(),
+                ],
+                res,
+            );
+            // 回退：撤销选择，恢复到之前的状态
+            undo_choice(state, choice.clone());
+        }
+    }
+}
+
+/* Driver Code */
+pub fn main() {
+    let root = vec_to_tree([1, 7, 3, 4, 5, 6, 7].map(|x| Some(x)).to_vec());
+    println!("初始化二叉树");
+    print_util::print_tree(root.as_ref().unwrap());
+
+    // 回溯算法
+    let mut res = Vec::new();
+    backtrack(&mut Vec::new(), &mut vec![root.unwrap()], &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 {
+        let mut vals = Vec::new();
+        for node in path {
+            vals.push(node.borrow().val)
+        }
+        println!("{:?}", vals);
+    }
+}
@@ -1,50 +1,56 @@
-/*
- * File: subset_sum_i.rs
- * Created Time: 2023-07-09
- * Author: codingonion (coderonion@gmail.com)
- */
-
-/* 回溯算法：子集和 I */
-fn backtrack(mut state: Vec<i32>, target: i32, choices: &[i32], start: usize, res: &mut Vec<Vec<i32>>) {
-    // 子集和等于 target 时，记录解
-    if target == 0 {
-        res.push(state);
-        return;
-    }
-    // 遍历所有选择
-    // 剪枝二：从 start 开始遍历，避免生成重复子集
-    for i in start..choices.len() {
-        // 剪枝一：若子集和超过 target ，则直接结束循环
-        // 这是因为数组已排序，后边元素更大，子集和一定超过 target
-        if target - choices[i] < 0 {
-            break;
-        }
-        // 尝试：做出选择，更新 target, start
-        state.push(choices[i]);
-        // 进行下一轮选择
-        backtrack(state.clone(), target - choices[i], choices, i, res);
-        // 回退：撤销选择，恢复到之前的状态
-        state.pop();
-    }
-}
-
-/* 求解子集和 I */
-fn subset_sum_i(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
-    let state = Vec::new(); // 状态（子集）
-    nums.sort(); // 对 nums 进行排序
-    let start = 0; // 遍历起始点
-    let mut res = Vec::new(); // 结果列表（子集列表）
-    backtrack(state, target, nums, start, &mut res);
-    res
-}
-
-/* Driver Code */
-pub fn main() {
-    let mut nums = [ 3, 4, 5 ];
-    let target = 9;
-
-    let res = subset_sum_i(&mut nums, target);
-    
-    println!("输入数组 nums = {:?}, target = {}", &nums, target);
-    println!("所有和等于 {} 的子集 res = {:?}", target, &res);
-}
+/*
+ * File: subset_sum_i.rs
+ * Created Time: 2023-07-09
+ * Author: codingonion (coderonion@gmail.com)
+ */
+
+/* 回溯算法：子集和 I */
+fn backtrack(
+    mut state: Vec<i32>,
+    target: i32,
+    choices: &[i32],
+    start: usize,
+    res: &mut Vec<Vec<i32>>,
+) {
+    // 子集和等于 target 时，记录解
+    if target == 0 {
+        res.push(state);
+        return;
+    }
+    // 遍历所有选择
+    // 剪枝二：从 start 开始遍历，避免生成重复子集
+    for i in start..choices.len() {
+        // 剪枝一：若子集和超过 target ，则直接结束循环
+        // 这是因为数组已排序，后边元素更大，子集和一定超过 target
+        if target - choices[i] < 0 {
+            break;
+        }
+        // 尝试：做出选择，更新 target, start
+        state.push(choices[i]);
+        // 进行下一轮选择
+        backtrack(state.clone(), target - choices[i], choices, i, res);
+        // 回退：撤销选择，恢复到之前的状态
+        state.pop();
+    }
+}
+
+/* 求解子集和 I */
+fn subset_sum_i(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
+    let state = Vec::new(); // 状态（子集）
+    nums.sort(); // 对 nums 进行排序
+    let start = 0; // 遍历起始点
+    let mut res = Vec::new(); // 结果列表（子集列表）
+    backtrack(state, target, nums, start, &mut res);
+    res
+}
+
+/* Driver Code */
+pub fn main() {
+    let mut nums = [3, 4, 5];
+    let target = 9;
+
+    let res = subset_sum_i(&mut nums, target);
+
+    println!("输入数组 nums = {:?}, target = {}", &nums, target);
+    println!("所有和等于 {} 的子集 res = {:?}", target, &res);
+}
@@ -1,48 +1,54 @@
-/*
- * File: subset_sum_i_naive.rs
- * Created Time: 2023-07-09
- * Author: codingonion (coderonion@gmail.com)
- */
-
-/* 回溯算法：子集和 I */
-fn backtrack(mut state: Vec<i32>, target: i32, total: i32, choices: &[i32], res: &mut Vec<Vec<i32>>) {
-    // 子集和等于 target 时，记录解
-    if total == target {
-        res.push(state);
-        return;
-    }
-    // 遍历所有选择
-    for i in 0..choices.len() {
-        // 剪枝：若子集和超过 target ，则跳过该选择
-        if total + choices[i] > target {
-            continue;
-        }
-        // 尝试：做出选择，更新元素和 total
-        state.push(choices[i]);
-        // 进行下一轮选择
-        backtrack(state.clone(), target, total + choices[i], choices, res);
-        // 回退：撤销选择，恢复到之前的状态
-        state.pop();
-    }
-}
-
-/* 求解子集和 I（包含重复子集） */
-fn subset_sum_i_naive(nums: &[i32], target: i32) -> Vec<Vec<i32>> {
-    let state = Vec::new(); // 状态（子集）
-    let total = 0; // 子集和
-    let mut res = Vec::new(); // 结果列表（子集列表）
-    backtrack(state, target, total, nums, &mut res);
-    res
-}
-
-/* Driver Code */
-pub fn main() {
-    let nums = [ 3, 4, 5 ];
-    let target = 9;
-
-    let res = subset_sum_i_naive(&nums, target);
-    
-    println!("输入数组 nums = {:?}, target = {}", &nums, target);
-    println!("所有和等于 {} 的子集 res = {:?}", target, &res);
-    println!("请注意，该方法输出的结果包含重复集合");
-}
+/*
+ * File: subset_sum_i_naive.rs
+ * Created Time: 2023-07-09
+ * Author: codingonion (coderonion@gmail.com)
+ */
+
+/* 回溯算法：子集和 I */
+fn backtrack(
+    mut state: Vec<i32>,
+    target: i32,
+    total: i32,
+    choices: &[i32],
+    res: &mut Vec<Vec<i32>>,
+) {
+    // 子集和等于 target 时，记录解
+    if total == target {
+        res.push(state);
+        return;
+    }
+    // 遍历所有选择
+    for i in 0..choices.len() {
+        // 剪枝：若子集和超过 target ，则跳过该选择
+        if total + choices[i] > target {
+            continue;
+        }
+        // 尝试：做出选择，更新元素和 total
+        state.push(choices[i]);
+        // 进行下一轮选择
+        backtrack(state.clone(), target, total + choices[i], choices, res);
+        // 回退：撤销选择，恢复到之前的状态
+        state.pop();
+    }
+}
+
+/* 求解子集和 I（包含重复子集） */
+fn subset_sum_i_naive(nums: &[i32], target: i32) -> Vec<Vec<i32>> {
+    let state = Vec::new(); // 状态（子集）
+    let total = 0; // 子集和
+    let mut res = Vec::new(); // 结果列表（子集列表）
+    backtrack(state, target, total, nums, &mut res);
+    res
+}
+
+/* Driver Code */
+pub fn main() {
+    let nums = [3, 4, 5];
+    let target = 9;
+
+    let res = subset_sum_i_naive(&nums, target);
+
+    println!("输入数组 nums = {:?}, target = {}", &nums, target);
+    println!("所有和等于 {} 的子集 res = {:?}", target, &res);
+    println!("请注意，该方法输出的结果包含重复集合");
+}
@@ -1,55 +1,61 @@
-/*
- * File: subset_sum_ii.rs
- * Created Time: 2023-07-09
- * Author: codingonion (coderonion@gmail.com)
- */
-
-/* 回溯算法：子集和 II */
-fn backtrack(mut state: Vec<i32>, target: i32, choices: &[i32], start: usize, res: &mut Vec<Vec<i32>>) {
-    // 子集和等于 target 时，记录解
-    if target == 0 {
-        res.push(state);
-        return;
-    }
-    // 遍历所有选择
-    // 剪枝二：从 start 开始遍历，避免生成重复子集
-    // 剪枝三：从 start 开始遍历，避免重复选择同一元素
-    for i in start..choices.len() {
-        // 剪枝一：若子集和超过 target ，则直接结束循环
-        // 这是因为数组已排序，后边元素更大，子集和一定超过 target
-        if target - choices[i] < 0 {
-            break;
-        }
-        // 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
-        if i > start && choices[i] == choices[i - 1] {
-            continue;
-        }
-        // 尝试：做出选择，更新 target, start
-        state.push(choices[i]);
-        // 进行下一轮选择
-        backtrack(state.clone(), target - choices[i], choices, i, res);
-        // 回退：撤销选择，恢复到之前的状态
-        state.pop();
-    }
-}
-
-/* 求解子集和 II */
-fn subset_sum_ii(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
-    let state = Vec::new(); // 状态（子集）
-    nums.sort(); // 对 nums 进行排序
-    let start = 0; // 遍历起始点
-    let mut res = Vec::new(); // 结果列表（子集列表）
-    backtrack(state, target, nums, start, &mut res);
-    res
-}
-
-/* Driver Code */
-pub fn main() {
-    let mut nums = [ 4, 4, 5 ];
-    let target = 9;
-
-    let res = subset_sum_ii(&mut nums, target);
-    
-    println!("输入数组 nums = {:?}, target = {}", &nums, target);
-    println!("所有和等于 {} 的子集 res = {:?}", target, &res);
-}
+/*
+ * File: subset_sum_ii.rs
+ * Created Time: 2023-07-09
+ * Author: codingonion (coderonion@gmail.com)
+ */
+
+/* 回溯算法：子集和 II */
+fn backtrack(
+    mut state: Vec<i32>,
+    target: i32,
+    choices: &[i32],
+    start: usize,
+    res: &mut Vec<Vec<i32>>,
+) {
+    // 子集和等于 target 时，记录解
+    if target == 0 {
+        res.push(state);
+        return;
+    }
+    // 遍历所有选择
+    // 剪枝二：从 start 开始遍历，避免生成重复子集
+    // 剪枝三：从 start 开始遍历，避免重复选择同一元素
+    for i in start..choices.len() {
+        // 剪枝一：若子集和超过 target ，则直接结束循环
+        // 这是因为数组已排序，后边元素更大，子集和一定超过 target
+        if target - choices[i] < 0 {
+            break;
+        }
+        // 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
+        if i > start && choices[i] == choices[i - 1] {
+            continue;
+        }
+        // 尝试：做出选择，更新 target, start
+        state.push(choices[i]);
+        // 进行下一轮选择
+        backtrack(state.clone(), target - choices[i], choices, i, res);
+        // 回退：撤销选择，恢复到之前的状态
+        state.pop();
+    }
+}
+
+/* 求解子集和 II */
+fn subset_sum_ii(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
+    let state = Vec::new(); // 状态（子集）
+    nums.sort(); // 对 nums 进行排序
+    let start = 0; // 遍历起始点
+    let mut res = Vec::new(); // 结果列表（子集列表）
+    backtrack(state, target, nums, start, &mut res);
+    res
+}
+
+/* Driver Code */
+pub fn main() {
+    let mut nums = [4, 4, 5];
+    let target = 9;
+
+    let res = subset_sum_ii(&mut nums, target);
+
+    println!("输入数组 nums = {:?}, target = {}", &nums, target);
+    println!("所有和等于 {} 的子集 res = {:?}", target, &res);
+}