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| 1 | +/** |
| 2 | + * @param {number} n |
| 3 | + * @return {string[][]} |
| 4 | + */ |
| 5 | +let solveNQueens = function (n) { |
| 6 | + let res = [] |
| 7 | + |
| 8 | + // 已摆放皇后的的列下标 |
| 9 | + let columns = [] |
| 10 | + // 已摆放皇后的对角线1下标 左下 -> 右上 |
| 11 | + // 计算某个坐标是否在这个对角线的方式是「行下标 + 列下标」是否相等 |
| 12 | + let dia1 = [] |
| 13 | + // 已摆放皇后的对角线2下标 左上 -> 右下 |
| 14 | + // 计算某个坐标是否在这个对角线的方式是「行下标 - 列下标」是否相等 |
| 15 | + let dia2 = [] |
| 16 | + |
| 17 | + // 尝试在一个n皇后问题中 摆放第index行内的皇后位置 |
| 18 | + let putQueen = (rowIndex, row) => { |
| 19 | + if (rowIndex === n) { |
| 20 | + res.push(generateBoard(row)) |
| 21 | + return |
| 22 | + } |
| 23 | + |
| 24 | + // 尝试摆第index行的皇后 尝试[0, n-1]列 |
| 25 | + for (let columnIndex = 0; columnIndex < n; columnIndex++) { |
| 26 | + // 在列上不冲突 |
| 27 | + let columnNotConflict = !columns[columnIndex] |
| 28 | + // 在对角线1上不冲突 |
| 29 | + let dia1NotConflict = !dia1[rowIndex + columnIndex] |
| 30 | + // 在对角线2上不冲突 |
| 31 | + let dia2NotConflict = !dia2[rowIndex - columnIndex] |
| 32 | + |
| 33 | + if (columnNotConflict && dia1NotConflict && dia2NotConflict) { |
| 34 | + |
| 35 | + columns[columnIndex] = true |
| 36 | + dia1[rowIndex + columnIndex] = true |
| 37 | + dia2[rowIndex - columnIndex] = true |
| 38 | + |
| 39 | + putQueen(rowIndex + 1, row.concat(columnIndex)) |
| 40 | + |
| 41 | + columns[columnIndex] = false |
| 42 | + dia1[rowIndex + columnIndex] = false |
| 43 | + dia2[rowIndex - columnIndex] = false |
| 44 | + } |
| 45 | + } |
| 46 | + } |
| 47 | + |
| 48 | + putQueen(0, []) |
| 49 | + |
| 50 | + return res |
| 51 | +} |
| 52 | + |
| 53 | +function generateBoard(row) { |
| 54 | + let n = row.length |
| 55 | + let res = [] |
| 56 | + for(let y = 0; y < n; y++) { |
| 57 | + let cur = '' |
| 58 | + for (let x = 0; x < n; x++) { |
| 59 | + if (x === row[y]) { |
| 60 | + cur += 'Q' |
| 61 | + }else { |
| 62 | + cur += '.' |
| 63 | + } |
| 64 | + } |
| 65 | + res.push(cur) |
| 66 | + } |
| 67 | + return res |
| 68 | +} |
| 69 | + |
| 70 | +console.log(solveNQueens(4)) |
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