feat: Add Dart codes for chapter_backtracking and chapter_divide_and_conquer (#680)

* feat: Add Dart codes for chapter_backtracking

* feat: Add Dart codes for chapter_divide_and_conquer

* Format Dart Code

codes/dart/chapter_backtracking/n_queens.dart

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+/**
+ * File: n_queens.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+/* 回溯算法：N 皇后 */
+void backtrack(
+  int row,
+  int n,
+  List<List<String>> state,
+  List<List<List<String>>> res,
+  List<bool> cols,
+  List<bool> diags1,
+  List<bool> diags2,
+) {
+  // 当放置完所有行时，记录解
+  if (row == n) {
+    List<List<String>> copyState = [];
+    for (List<String> sRow in state) {
+      copyState.add(List.from(sRow));
+    }
+    res.add(copyState);
+    return;
+  }
+  // 遍历所有列
+  for (int col = 0; col < n; col++) {
+    // 计算该格子对应的主对角线和副对角线
+    int diag1 = row - col + n - 1;
+    int 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[row][col] = "Q";
+      cols[col] = true;
+      diags1[diag1] = true;
+      diags2[diag2] = 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[row][col] = "#";
+      cols[col] = false;
+      diags1[diag1] = false;
+      diags2[diag2] = false;
+    }
+  }
+}
+
+/* 求解 N 皇后 */
+List<List<List<String>>> nQueens(int n) {
+  // 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
+  List<List<String>> state = List.generate(n, (index) => List.filled(n, "#"));
+  List<bool> cols = List.filled(n, false); // 记录列是否有皇后
+  List<bool> diags1 = List.filled(2 * n - 1, false); // 记录主对角线是否有皇后
+  List<bool> diags2 = List.filled(2 * n - 1, false); // 记录副对角线是否有皇后
+  List<List<List<String>>> res = [];
+
+  backtrack(0, n, state, res, cols, diags1, diags2);
+
+  return res;
+}
+
+/* Driver Code */
+void main() {
+  int n = 4;
+  List<List<List<String>>> res = nQueens(n);
+  print("输入棋盘长宽为 $n");
+  print("皇后放置方案共有 ${res.length} 种");
+  for (List<List<String>> state in res) {
+    print("--------------------");
+    for (List<String> row in state) {
+      print(row);
+    }
+  }
+}

codes/dart/chapter_backtracking/permutations_i.dart

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+/**
+ * File: permutations_i.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+/* 回溯算法：全排列 I */
+void backtrack(
+  List<int> state,
+  List<int> choices,
+  List<bool> selected,
+  List<List<int>> res,
+) {
+  // 当状态长度等于元素数量时，记录解
+  if (state.length == choices.length) {
+    res.add(List.from(state));
+    return;
+  }
+  // 遍历所有选择
+  for (int i = 0; i < choices.length; i++) {
+    int 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.add(choice);
+      // 进行下一轮选择
+      backtrack(state, choices, selected, res);
+      // 回退：撤销选择，恢复到之前的状态
+      selected[i] = false;
+      state.removeLast();
+    }
+  }
+}
+
+/* 全排列 I */
+List<List<int>> permutationsI(List<int> nums) {
+  List<List<int>> res = [];
+  backtrack([], nums, List.filled(nums.length, false), res);
+  return res;
+}
+
+/* Driver Code */
+void main() {
+  List<int> nums = [1, 2, 3];
+
+  List<List<int>> res = permutationsI(nums);
+
+  print("输入数组 nums = $nums");
+  print("所有排列 res = $res");
+}

codes/dart/chapter_backtracking/permutations_ii.dart

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+/**
+ * File: permutations_ii.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+/* 回溯算法：全排列 II */
+void backtrack(
+  List<int> state,
+  List<int> choices,
+  List<bool> selected,
+  List<List<int>> res,
+) {
+  // 当状态长度等于元素数量时，记录解
+  if (state.length == choices.length) {
+    res.add(List.from(state));
+    return;
+  }
+  // 遍历所有选择
+  Set<int> duplicated = {};
+  for (int i = 0; i < choices.length; i++) {
+    int 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.add(choice); // 记录选择过的元素值
+      selected[i] = true;
+      state.add(choice);
+      // 进行下一轮选择
+      backtrack(state, choices, selected, res);
+      // 回退：撤销选择，恢复到之前的状态
+      selected[i] = false;
+      state.removeLast();
+    }
+  }
+}
+
+/* 全排列 II */
+List<List<int>> permutationsII(List<int> nums) {
+  List<List<int>> res = [];
+  backtrack([], nums, List.filled(nums.length, false), res);
+  return res;
+}
+
+/* Driver Code */
+void main() {
+  List<int> nums = [1, 2, 2];
+
+  List<List<int>> res = permutationsII(nums);
+
+  print("输入数组 nums = $nums");
+  print("所有排列 res = $res");
+}

codes/dart/chapter_backtracking/preorder_traversal_i_compact.dart

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+/**
+ * File: preorder_traversal_i_compact.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+import '../utils/print_util.dart';
+import '../utils/tree_node.dart';
+
+/* 前序遍历：例题一 */
+void preOrder(TreeNode? root, List<TreeNode> res) {
+  if (root == null) {
+    return;
+  }
+  if (root.val == 7) {
+    // 记录解
+    res.add(root);
+  }
+  preOrder(root.left, res);
+  preOrder(root.right, res);
+}
+
+/* Driver Code */
+void main() {
+  TreeNode? root = listToTree([1, 7, 3, 4, 5, 6, 7]);
+  print("\n初始化二叉树");
+  printTree(root);
+
+  // 前序遍历
+  List<TreeNode> res = [];
+  preOrder(root, res);
+
+  print("\n输出所有值为 7 的节点");
+  print(List.generate(res.length, (i) => res[i].val));
+}

codes/dart/chapter_backtracking/preorder_traversal_ii_compact.dart

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+/**
+ * File: preorder_traversal_ii_compact.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+import '../utils/print_util.dart';
+import '../utils/tree_node.dart';
+
+/* 前序遍历：例题二 */
+void preOrder(
+  TreeNode? root,
+  List<TreeNode> path,
+  List<List<TreeNode>> res,
+) {
+  if (root == null) {
+    return;
+  }
+
+  // 尝试
+  path.add(root);
+  if (root.val == 7) {
+    // 记录解
+    res.add(List.from(path));
+  }
+  preOrder(root.left, path, res);
+  preOrder(root.right, path, res);
+  // 回退
+  path.removeLast();
+}
+
+/* Driver Code */
+void main() {
+  TreeNode? root = listToTree([1, 7, 3, 4, 5, 6, 7]);
+  print("\n初始化二叉树");
+  printTree(root);
+
+  // 前序遍历
+  List<TreeNode> path = [];
+  List<List<TreeNode>> res = [];
+  preOrder(root, path, res);
+
+  print("\n输出所有根节点到节点 7 的路径");
+  for (List<TreeNode> vals in res) {
+    print(List.generate(vals.length, (i) => vals[i].val));
+  }
+}

codes/dart/chapter_backtracking/preorder_traversal_iii_compact.dart

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+/**
+ * File: preorder_traversal_iii_compact.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+import '../utils/print_util.dart';
+import '../utils/tree_node.dart';
+
+/* 前序遍历：例题三 */
+void preOrder(
+  TreeNode? root,
+  List<TreeNode> path,
+  List<List<TreeNode>> res,
+) {
+  if (root == null || root.val == 3) {
+    return;
+  }
+
+  // 尝试
+  path.add(root);
+  if (root.val == 7) {
+    // 记录解
+    res.add(List.from(path));
+    path.removeLast();
+    return;
+  }
+  preOrder(root.left, path, res);
+  preOrder(root.right, path, res);
+  // 回退
+  path.removeLast();
+}
+
+/* Driver Code */
+void main() {
+  TreeNode? root = listToTree([1, 7, 3, 4, 5, 6, 7]);
+  print("\n初始化二叉树");
+  printTree(root);
+
+  // 前序遍历
+  List<TreeNode> path = [];
+  List<List<TreeNode>> res = [];
+  preOrder(root, path, res);
+
+  print("\n输出所有根节点到节点 7 的路径");
+  for (List<TreeNode> vals in res) {
+    print(List.generate(vals.length, (i) => vals[i].val));
+  }
+}

codes/dart/chapter_backtracking/preorder_traversal_iii_template.dart

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+/**
+ * File: preorder_traversal_iii_template.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+import '../utils/print_util.dart';
+import '../utils/tree_node.dart';
+
+/* 判断当前状态是否为解 */
+bool isSolution(List<TreeNode> state) {
+  return state.isNotEmpty && state.last.val == 7;
+}
+
+/* 记录解 */
+void recordSolution(List<TreeNode> state, List<List<TreeNode>> res) {
+  res.add(List.from(state));
+}
+
+/* 判断在当前状态下，该选择是否合法 */
+bool isValid(List<TreeNode> state, TreeNode? choice) {
+  return choice != null && choice.val != 3;
+}
+
+/* 更新状态 */
+void makeChoice(List<TreeNode> state, TreeNode? choice) {
+  state.add(choice!);
+}
+
+/* 恢复状态 */
+void undoChoice(List<TreeNode> state, TreeNode? choice) {
+  state.removeLast();
+}
+
+/* 回溯算法：例题三 */
+void backtrack(
+  List<TreeNode> state,
+  List<TreeNode?> choices,
+  List<List<TreeNode>> res,
+) {
+  // 检查是否为解
+  if (isSolution(state)) {
+    // 记录解
+    recordSolution(state, res);
+  }
+  // 遍历所有选择
+  for (TreeNode? choice in choices) {
+    // 剪枝：检查选择是否合法
+    if (isValid(state, choice)) {
+      // 尝试：做出选择，更新状态
+      makeChoice(state, choice);
+      // 进行下一轮选择
+      backtrack(state, [choice!.left, choice.right], res);
+      // 回退：撤销选择，恢复到之前的状态
+      undoChoice(state, choice);
+    }
+  }
+}
+
+/* Driver Code */
+void main() {
+  TreeNode? root = listToTree([1, 7, 3, 4, 5, 6, 7]);
+  print("\n初始化二叉树");
+  printTree(root);
+
+  // 回溯算法
+  List<List<TreeNode>> res = [];
+  backtrack([], [root!], res);
+  print("\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 (List<TreeNode> path in res) {
+    print(List.from(path.map((e) => e.val)));
+  }
+}

codes/dart/chapter_backtracking/subset_sum_i.dart

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+/**
+ * File: subset_sum_i.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+/* 回溯算法：子集和 I */
+void backtrack(
+  List<int> state,
+  int target,
+  List<int> choices,
+  int start,
+  List<List<int>> res,
+) {
+  // 子集和等于 target 时，记录解
+  if (target == 0) {
+    res.add(List.from(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 (int i = start; i < choices.length; i++) {
+    // 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.add(choices[i]);
+    // 进行下一轮选择
+    backtrack(state, target - choices[i], choices, i, res);
+    // 回退：撤销选择，恢复到之前的状态
+    state.removeLast();
+  }
+}
+
+/* 求解子集和 I */
+List<List<int>> subsetSumI(List<int> nums, int target) {
+  List<int> state = []; // 状态（子集）
+  nums.sort(); // 对 nums 进行排序
+  int start = 0; // 遍历起始点
+  List<List<int>> res = []; // 结果列表（子集列表）
+  backtrack(state, target, nums, start, res);
+  return res;
+}
+
+/* Driver Code */
+void main() {
+  List<int> nums = [3, 4, 5];
+  int target = 9;
+
+  List<List<int>> res = subsetSumI(nums, target);
+
+  print("输入数组 nums = $nums, target = $target");
+  print("所有和等于 $target 的子集 res = $res");
+}

codes/dart/chapter_backtracking/subset_sum_i_naive.dart

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+/**
+ * File: subset_sum_i_naive.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+/* 回溯算法：子集和 I */
+void backtrack(
+  List<int> state,
+  int target,
+  int total,
+  List<int> choices,
+  List<List<int>> res,
+) {
+  // 子集和等于 target 时，记录解
+  if (total == target) {
+    res.add(List.from(state));
+    return;
+  }
+  // 遍历所有选择
+  for (int i = 0; i < choices.length; i++) {
+    // 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.add(choices[i]);
+    // 进行下一轮选择
+    backtrack(state, target, total + choices[i], choices, res);
+    // 回退：撤销选择，恢复到之前的状态
+    state.removeLast();
+  }
+}
+
+/* 求解子集和 I（包含重复子集） */
+List<List<int>> subsetSumINaive(List<int> nums, int target) {
+  List<int> state = []; // 状态（子集）
+  int total = 0; // 元素和
+  List<List<int>> res = []; // 结果列表（子集列表）
+  backtrack(state, target, total, nums, res);
+  return res;
+}
+
+/* Driver Code */
+void main() {
+  List<int> nums = [3, 4, 5];
+  int target = 9;
+
+  List<List<int>> res = subsetSumINaive(nums, target);
+
+  print("输入数组 nums = $nums, target = $target");
+  print("所有和等于 $target 的子集 res = $res");
+  print("请注意，该方法输出的结果包含重复集合");
+}

codes/dart/chapter_backtracking/subset_sum_ii.dart

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+/**
+ * File: subset_sum_ii.dart
+ * Created Time: 2023-08-10
+ * Author: liuyuxin (gvenusleo@gmail.com)
+ */
+
+/* 回溯算法：子集和 II */
+void backtrack(
+  List<int> state,
+  int target,
+  List<int> choices,
+  int start,
+  List<List<int>> res,
+) {
+  // 子集和等于 target 时，记录解
+  if (target == 0) {
+    res.add(List.from(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 (int i = start; i < choices.length; i++) {
+    // 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.add(choices[i]);
+    // 进行下一轮选择
+    backtrack(state, target - choices[i], choices, i + 1, res);
+    // 回退：撤销选择，恢复到之前的状态
+    state.removeLast();
+  }
+}
+
+/* 求解子集和 II */
+List<List<int>> subsetSumII(List<int> nums, int target) {
+  List<int> state = []; // 状态（子集）
+  nums.sort(); // 对 nums 进行排序
+  int start = 0; // 遍历起始点
+  List<List<int>> res = []; // 结果列表（子集列表）
+  backtrack(state, target, nums, start, res);
+  return res;
+}
+
+void main() {
+  List<int> nums = [4, 4, 5];
+  int target = 9;
+
+  List<List<int>> res = subsetSumII(nums, target);
+
+  print("输入数组 nums = $nums, target = $target");
+  print("所有和等于 $target 的子集 res = $res");
+}