gold-8-12


设计一种算法,打印 N 皇后在 N × N 棋盘上的各种摆法,其中每个皇后都不同行、不同列,也不在对角线上。这里的“对角线”指的是所有的对角线,不只是平分整个棋盘的那两条对角线。

注意:本题相对原题做了扩展

示例:

 输入:4
 输出:[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
 解释: 4 皇后问题存在如下两个不同的解法。
[
 [".Q..",  // 解法 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // 解法 2
  "Q...",
  "...Q",
  ".Q.."]
]
class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> solutions = new ArrayList<List<String>>();
        int[] queens = new int[n];
        Arrays.fill(queens, -1);
        Set<Integer> columns = new HashSet<Integer>();
        Set<Integer> diagonals1 = new HashSet<Integer>();
        Set<Integer> diagonals2 = new HashSet<Integer>();
        backtrack(solutions, queens, n, 0, columns, diagonals1, diagonals2);
        return solutions;
    }

    public void backtrack(List<List<String>> solutions, int[] queens, int n, int row, Set<Integer> columns, Set<Integer> diagonals1, Set<Integer> diagonals2) {
        if (row == n) {
            List<String> board = generateBoard(queens, n);
            solutions.add(board);
        } else {
            for (int i = 0; i < n; i++) {
                if (columns.contains(i)) {
                    continue;
                }
                int diagonal1 = row - i;
                if (diagonals1.contains(diagonal1)) {
                    continue;
                }
                int diagonal2 = row + i;
                if (diagonals2.contains(diagonal2)) {
                    continue;
                }
                queens[row] = i;
                columns.add(i);
                diagonals1.add(diagonal1);
                diagonals2.add(diagonal2);
                backtrack(solutions, queens, n, row + 1, columns, diagonals1, diagonals2);
                queens[row] = -1;
                columns.remove(i);
                diagonals1.remove(diagonal1);
                diagonals2.remove(diagonal2);
            }
        }
    }

    public List<String> generateBoard(int[] queens, int n) {
        List<String> board = new ArrayList<String>();
        for (int i = 0; i < n; i++) {
            char[] row = new char[n];
            Arrays.fill(row, '.');
            row[queens[i]] = 'Q';
            board.add(new String(row));
        }
        return board;
    }
}

文章作者: 倪春恩
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