Ref: https://pdai.tech/md/algorithm/alg-core-backtracking.html#n-%E7%9A%87%E5%90%8E
在 n*n 的矩阵中摆放 n 个皇后,并且每个皇后不能在同一行,同一列,同一对角线上,求所有的 n 皇后的解。
一行一行地摆放,在确定一行中的那个皇后应该摆在哪一列时,需要用三个标记数组来确定某一列是否合法,这三个标记数组分别为:列标记数组、45 度对角线标记数组和 135 度对角线标记数组。
45 度对角线标记数组的维度为 2 n - 1,通过下图可以明确 (r, c) 的位置所在的数组下标为 r + c:
135 度对角线标记数组的维度也是 2 n - 1,(r, c) 的位置所在的数组下标为 n - 1 - (r - c):
private List<List<String>> solutions;
private char[][] nQueens;
private boolean[] colUsed;
private boolean[] diagonals45Used;
private boolean[] diagonals135Used;
private int n;
public List<List<String>> solveNQueens(int n) {
solutions = new ArrayList<>();
nQueens = new char[n][n];
for (int i = 0; i < n; i++) {
Arrays.fill(nQueens[i], '.');
}
colUsed = new boolean[n];
diagonals45Used = new boolean[2 * n - 1];
diagonals135Used = new boolean[2 * n - 1];
this.n = n;
backtracking(0);
return solutions;
}
private void backtracking(int row) {
if (row == n) {
List<String> list = new ArrayList<>();
for (char[] chars : nQueens) {
list.add(new String(chars));
}
solutions.add(list);
return;
}
for (int col = 0; col < n; col++) {
int diagonals45Idx = row + col;
int diagonals135Idx = n - 1 - (row - col);
if (colUsed[col] || diagonals45Used[diagonals45Idx] || diagonals135Used[diagonals135Idx]) {
continue;
}
nQueens[row][col] = 'Q';
colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = true;
backtracking(row + 1);
colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = false;
nQueens[row][col] = '.';
}
}