rust and zig : add codes for chapter_dynamic_programming (#606)

* rust : add codes for chapter_dynamic_programming

* zig : add codes for chapter_dynamic_programming

* rust : add codes for chapter_backtracking

* Update n_queens.rs

---------

Co-authored-by: Yudong Jin <krahets@163.com>

codes/rust/chapter_backtracking/n_queens.rs

@@ -0,0 +1,72 @@
+/*
+ * File: n_queens.rs
+ * Created Time: 2023-07-15
+ * Author: sjinzh (sjinzh@gmail.com)
+ */
+
+/* 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]) {
+    // 当放置完所有行时，记录解
+    if row == n {
+        let mut copy_state: Vec<Vec<String>> = Vec::new();
+        for s_row in state.clone() {
+            copy_state.push(s_row);
+        }
+        res.push(copy_state);
+        return;
+    }
+    // 遍历所有列
+    for col in 0..n {
+        // 计算该格子对应的主对角线和副对角线
+        let diag1 = row + n - 1 - col;
+        let diag2 = row + col;
+        // 剪枝：不允许该格子所在列、主对角线、副对角线存在皇后
+        if !cols[col] && !diags1[diag1] && !diags2[diag2] {
+            // 尝试：将皇后放置在该格子
+            state.get_mut(row).unwrap()[col] = "Q".into();
+            (cols[col], diags1[diag1], diags2[diag2]) = (true, true, true);
+            // 放置下一行
+            backtrack(row + 1, n, state, res, cols, diags1, diags2);
+            // 回退：将该格子恢复为空位
+            state.get_mut(row).unwrap()[col] = "#".into();
+            (cols[col], diags1[diag1], diags2[diag2]) = (false, false, false);
+        }
+    }
+}
+
+/* 求解 N 皇后 */
+fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
+    // 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
+    let mut state: Vec<Vec<String>> = Vec::new();
+    for _ in 0..n {
+        let mut row: Vec<String> = Vec::new();
+        for _ in 0..n {
+            row.push("#".into());
+        }
+        state.push(row);
+    }
+    let mut cols = vec![false; n]; // 记录列是否有皇后
+    let mut diags1 = vec![false; 2 * n - 1]; // 记录主对角线是否有皇后
+    let mut diags2 = vec![false; 2 * n - 1]; // 记录副对角线是否有皇后
+    let mut res: Vec<Vec<Vec<String>>> = Vec::new();
+
+    backtrack(0, n, &mut state, &mut res, &mut cols, &mut diags1, &mut diags2);
+
+    res
+}
+
+/* Driver Code */
+pub fn main() {
+    let n: usize = 4;
+    let res = n_queens(n);
+
+    println!("输入棋盘长宽为 {n}");
+    println!("皇后放置方案共有 {} 种", res.len());
+    for state in res.iter() {
+        println!("--------------------");
+        for row in state.iter() {
+            println!("{:?}", row);
+        }
+    }
+}

codes/rust/chapter_backtracking/permutations_i.rs

@@ -0,0 +1,46 @@
+/*
+ * File: permutations_i.rs
+ * Created Time: 2023-07-15
+ * Author: sjinzh (sjinzh@gmail.com)
+ */
+
+/* 回溯算法：全排列 I */
+fn backtrack(mut state: Vec<i32>, choices: &[i32], selected: &mut [bool], res: &mut Vec<Vec<i32>>) {
+    // 当状态长度等于元素数量时，记录解
+    if state.len() == choices.len() {
+        res.push(state);
+        return;
+    }
+    // 遍历所有选择
+    for i in 0..choices.len() {
+        let choice = choices[i];
+        // 剪枝：不允许重复选择元素 且 不允许重复选择相等元素
+        if !selected[i] {
+            // 尝试：做出选择，更新状态
+            selected[i] = true;
+            state.push(choice);
+            // 进行下一轮选择
+            backtrack(state.clone(), choices, selected, res);
+            // 回退：撤销选择，恢复到之前的状态
+            selected[i] = false;
+            state.remove(state.len() - 1);
+        }
+    }
+}
+
+/* 全排列 I */
+fn permutations_i(nums: &mut [i32]) -> Vec<Vec<i32>> {
+    let mut res = Vec::new(); // 状态（子集）
+    backtrack(Vec::new(), nums, &mut vec![false; nums.len()], &mut res);
+    res
+}
+
+/* Driver Code */
+pub fn main() {
+    let mut nums = [ 1, 2, 3 ];
+
+    let res = permutations_i(&mut nums);
+
+    println!("输入数组 nums = {:?}", &nums);
+    println!("所有排列 res = {:?}", &res);
+}

codes/rust/chapter_backtracking/permutations_ii.rs

@@ -0,0 +1,50 @@
+/*
+ * File: permutations_ii.rs
+ * Created Time: 2023-07-15
+ * Author: sjinzh (sjinzh@gmail.com)
+ */
+
+use std::collections::HashSet;
+
+/* 回溯算法：全排列 II */
+fn backtrack(mut state: Vec<i32>, choices: &[i32], selected: &mut [bool], res: &mut Vec<Vec<i32>>) {
+    // 当状态长度等于元素数量时，记录解
+    if state.len() == choices.len() {
+        res.push(state);
+        return;
+    }
+    // 遍历所有选择
+    let mut duplicated = HashSet::<i32>::new();
+    for i in 0..choices.len() {
+        let choice = choices[i];
+        // 剪枝：不允许重复选择元素 且 不允许重复选择相等元素
+        if !selected[i] && !duplicated.contains(&choice) {
+            // 尝试：做出选择，更新状态
+            duplicated.insert(choice); // 记录选择过的元素值
+            selected[i] = true;
+            state.push(choice);
+            // 进行下一轮选择
+            backtrack(state.clone(), choices, selected, res);
+            // 回退：撤销选择，恢复到之前的状态
+            selected[i] = false;
+            state.remove(state.len() - 1);
+        }
+    }
+}
+
+/* 全排列 II */
+fn permutations_ii(nums: &mut [i32]) -> Vec<Vec<i32>> {
+    let mut res = Vec::new();
+    backtrack(Vec::new(), nums, &mut vec![false; nums.len()], &mut res);
+    res
+}
+
+/* Driver Code */
+pub fn main() {
+    let mut nums = [ 1, 2, 2 ];
+
+    let res = permutations_ii(&mut nums);
+
+    println!("输入数组 nums = {:?}", &nums);
+    println!("所有排列 res = {:?}", &res);
+}