Add JavaScript and TypeScript code of permutations and n_queens (Chap…

…ter of Backtracking) (#494) * Add JavaScript and TypeScript code of permutations and n_queens (Chapter of Backtracking) * Update n_queens.js * Update permutations_i.js * Update permutations_ii.js * Update n_queens.ts * Update permutations_i.ts * Update permutations_ii.ts --------- Co-authored-by: Yudong Jin <krahets@163.com>
krahets · May 13, 2023 · b52a98f · b52a98f
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diff --git a/codes/javascript/chapter_backtracking/n_queens.js b/codes/javascript/chapter_backtracking/n_queens.js
@@ -0,0 +1,55 @@
+/**
+ * File: n_queens.js
+ * Created Time: 2023-05-13
+ * Author: Justin (xiefahit@gmail.com)
+ */
+
+/* 回溯算法：N 皇后 */
+function backtrack(row, n, state, res, cols, diags1, diags2) {
+    // 当放置完所有行时，记录解
+    if (row === n) {
+        res.push(state.map((row) => row.slice()));
+        return;
+    }
+    // 遍历所有列
+    for (let col = 0; col < n; col++) {
+        // 计算该格子对应的主对角线和副对角线
+        const diag1 = row - col + n - 1;
+        const diag2 = row + col;
+        // 剪枝：不允许该格子所在 (列 或 主对角线 或 副对角线) 包含皇后
+        if (!(cols[col] || diags1[diag1] || diags2[diag2])) {
+            // 尝试：将皇后放置在该格子
+            state[row][col] = 'Q';
+            cols[col] = diags1[diag1] = 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] = diags1[diag1] = diags2[diag2] = false;
+        }
+    }
+}
+
+/* 求解 N 皇后 */
+function nQueens(n) {
+    // 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
+    const state = Array.from({ length: n }, () => Array(n).fill('#'));
+    const cols = Array(n).fill(false); // 记录列是否有皇后
+    const diags1 = Array(2 * n - 1).fill(false); // 记录主对角线是否有皇后
+    const diags2 = Array(2 * n - 1).fill(false); // 记录副对角线是否有皇后
+    const res = [];
+
+    backtrack(0, n, state, res, cols, diags1, diags2);
+    return res;
+}
+
+// Driver Code
+const n = 4;
+const res = nQueens(n);
+
+console.log(`输入棋盘长宽为 ${n}`);
+console.log(`皇后放置方案共有 ${res.length} 种`);
+res.forEach((state) => {
+    console.log('--------------------');
+    state.forEach((row) => console.log(row));
+});
diff --git a/codes/javascript/chapter_backtracking/permutations_i.js b/codes/javascript/chapter_backtracking/permutations_i.js
@@ -0,0 +1,42 @@
+/**
+ * File: permutations_i.js
+ * Created Time: 2023-05-13
+ * Author: Justin (xiefahit@gmail.com)
+ */
+
+/* 回溯算法：全排列 I */
+function backtrack(state, choices, selected, res) {
+    // 当状态长度等于元素数量时，记录解
+    if (state.length === choices.length) {
+        res.push([...state]);
+        return;
+    }
+    // 遍历所有选择
+    choices.forEach((choice, 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, choices, selected, res);
+            // 回退：撤销选择，恢复到之前的状态
+            selected[i] = false;
+            state.pop();
+        }
+    });
+}
+
+/* 全排列 I */
+function permutationsI(nums) {
+    const res = [];
+    backtrack([], nums, Array(nums.length).fill(false), res);
+    return res;
+}
+
+// Driver Code
+const nums = [1, 2, 3];
+const res = permutationsI(nums);
+
+console.log(`输入数组 nums = ${JSON.stringify(nums)}`);
+console.log(`所有排列 res = ${JSON.stringify(res)}`);
diff --git a/codes/javascript/chapter_backtracking/permutations_ii.js b/codes/javascript/chapter_backtracking/permutations_ii.js
@@ -0,0 +1,44 @@
+/**
+ * File: permutations_ii.js
+ * Created Time: 2023-05-13
+ * Author: Justin (xiefahit@gmail.com)
+ */
+
+/* 回溯算法：全排列 II */
+function backtrack(state, choices, selected, res) {
+    // 当状态长度等于元素数量时，记录解
+    if (state.length === choices.length) {
+        res.push([...state]);
+        return;
+    }
+    // 遍历所有选择
+    const duplicated = new Set();
+    choices.forEach((choice, 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.has(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.push(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.pop();
+        }
+    });
+}
+
+/* 全排列 II */
+function permutationsII(nums) {
+    const res = [];
+    backtrack([], nums, Array(nums.length).fill(false), res);
+    return res;
+}
+
+// Driver Code
+const nums = [1, 2, 2];
+const res = permutationsII(nums);
+
+console.log(`输入数组 nums = ${JSON.stringify(nums)}`);
+console.log(`所有排列 res = ${JSON.stringify(res)}`);
diff --git a/codes/typescript/chapter_backtracking/n_queens.ts b/codes/typescript/chapter_backtracking/n_queens.ts
@@ -0,0 +1,65 @@
+/**
+ * File: n_queens.ts
+ * Created Time: 2023-05-13
+ * Author: Justin (xiefahit@gmail.com)
+ */
+
+/* 回溯算法：N 皇后 */
+function backtrack(
+    row: number,
+    n: number,
+    state: string[][],
+    res: string[][][],
+    cols: boolean[],
+    diags1: boolean[],
+    diags2: boolean[]
+): void {
+    // 当放置完所有行时，记录解
+    if (row === n) {
+        res.push(state.map((row) => row.slice()));
+        return;
+    }
+    // 遍历所有列
+    for (let col = 0; col < n; col++) {
+        // 计算该格子对应的主对角线和副对角线
+        const diag1 = row - col + n - 1;
+        const diag2 = row + col;
+        // 剪枝：不允许该格子所在 (列 或 主对角线 或 副对角线) 包含皇后
+        if (!(cols[col] || diags1[diag1] || diags2[diag2])) {
+            // 尝试：将皇后放置在该格子
+            state[row][col] = 'Q';
+            cols[col] = diags1[diag1] = 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] = diags1[diag1] = diags2[diag2] = false;
+        }
+    }
+}
+
+/* 求解 N 皇后 */
+function nQueens(n: number): string[][][] {
+    // 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
+    const state = Array.from({ length: n }, () => Array(n).fill('#'));
+    const cols = Array(n).fill(false); // 记录列是否有皇后
+    const diags1 = Array(2 * n - 1).fill(false); // 记录主对角线是否有皇后
+    const diags2 = Array(2 * n - 1).fill(false); // 记录副对角线是否有皇后
+    const res: string[][][] = [];
+
+    backtrack(0, n, state, res, cols, diags1, diags2);
+    return res;
+}
+
+// Driver Code
+const n = 4;
+const res = nQueens(n);
+
+console.log(`输入棋盘长宽为 ${n}`);
+console.log(`皇后放置方案共有 ${res.length} 种`);
+res.forEach((state) => {
+    console.log('--------------------');
+    state.forEach((row) => console.log(row));
+});
+
+export {};
diff --git a/codes/typescript/chapter_backtracking/permutations_i.ts b/codes/typescript/chapter_backtracking/permutations_i.ts
@@ -0,0 +1,49 @@
+/**
+ * File: permutations_i.ts
+ * Created Time: 2023-05-13
+ * Author: Justin (xiefahit@gmail.com)
+ */
+
+/* 回溯算法：全排列 I */
+function backtrack(
+    state: number[],
+    choices: number[],
+    selected: boolean[],
+    res: number[][]
+): void {
+    // 当状态长度等于元素数量时，记录解
+    if (state.length === choices.length) {
+        res.push([...state]);
+        return;
+    }
+    // 遍历所有选择
+    choices.forEach((choice, 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, choices, selected, res);
+            // 回退：撤销选择，恢复到之前的状态
+            selected[i] = false;
+            state.pop();
+        }
+    });
+}
+
+/* 全排列 I */
+function permutationsI(nums: number[]): number[][] {
+    const res: number[][] = [];
+    backtrack([], nums, Array(nums.length).fill(false), res);
+    return res;
+}
+
+// Driver Code
+const nums: number[] = [1, 2, 3];
+const res: number[][] = permutationsI(nums);
+
+console.log(`输入数组 nums = ${JSON.stringify(nums)}`);
+console.log(`所有排列 res = ${JSON.stringify(res)}`);
+
+export {};
diff --git a/codes/typescript/chapter_backtracking/permutations_ii.ts b/codes/typescript/chapter_backtracking/permutations_ii.ts
@@ -0,0 +1,51 @@
+/**
+ * File: permutations_ii.ts
+ * Created Time: 2023-05-13
+ * Author: Justin (xiefahit@gmail.com)
+ */
+
+/* 回溯算法：全排列 II */
+function backtrack(
+    state: number[],
+    choices: number[],
+    selected: boolean[],
+    res: number[][]
+): void {
+    // 当状态长度等于元素数量时，记录解
+    if (state.length === choices.length) {
+        res.push([...state]);
+        return;
+    }
+    // 遍历所有选择
+    const duplicated = new Set();
+    choices.forEach((choice, 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.has(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.push(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.pop();
+        }
+    });
+}
+
+/* 全排列 II */
+function permutationsII(nums: number[]): number[][] {
+    const res: number[][] = [];
+    backtrack([], nums, Array(nums.length).fill(false), res);
+    return res;
+}
+
+// Driver Code
+const nums: number[] = [1, 2, 2];
+const res: number[][] = permutationsII(nums);
+
+console.log(`输入数组 nums = ${JSON.stringify(nums)}`);
+console.log(`所有排列 res = ${JSON.stringify(res)}`);
+
+export {};