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* 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>
55 lines
1.9 KiB
JavaScript
55 lines
1.9 KiB
JavaScript
/**
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* File: n_queens.js
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* Created Time: 2023-05-13
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* Author: Justin (xiefahit@gmail.com)
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*/
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/* 回溯算法:N 皇后 */
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function backtrack(row, n, state, res, cols, diags1, diags2) {
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// 当放置完所有行时,记录解
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if (row === n) {
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res.push(state.map((row) => row.slice()));
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return;
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}
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// 遍历所有列
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for (let col = 0; col < n; col++) {
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// 计算该格子对应的主对角线和副对角线
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const diag1 = row - col + n - 1;
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const diag2 = row + col;
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// 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
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if (!(cols[col] || diags1[diag1] || diags2[diag2])) {
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// 尝试:将皇后放置在该格子
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state[row][col] = 'Q';
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cols[col] = diags1[diag1] = diags2[diag2] = true;
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// 放置下一行
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backtrack(row + 1, n, state, res, cols, diags1, diags2);
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// 回退:将该格子恢复为空位
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state[row][col] = '#';
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cols[col] = diags1[diag1] = diags2[diag2] = false;
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}
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}
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}
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/* 求解 N 皇后 */
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function nQueens(n) {
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// 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位
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const state = Array.from({ length: n }, () => Array(n).fill('#'));
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const cols = Array(n).fill(false); // 记录列是否有皇后
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const diags1 = Array(2 * n - 1).fill(false); // 记录主对角线是否有皇后
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const diags2 = Array(2 * n - 1).fill(false); // 记录副对角线是否有皇后
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const res = [];
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backtrack(0, n, state, res, cols, diags1, diags2);
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return res;
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}
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// Driver Code
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const n = 4;
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const res = nQueens(n);
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console.log(`输入棋盘长宽为 ${n}`);
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console.log(`皇后放置方案共有 ${res.length} 种`);
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res.forEach((state) => {
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console.log('--------------------');
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state.forEach((row) => console.log(row));
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});
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