[LeetCode] 51. N-Queens N皇后问题

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

Example:

Input: 4
Output: [
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

八皇后问题扩展到N皇后，经典的回溯算法题。

解法：回溯Backtracking

Java:

public class Solution {
    public List<List<String>> solveNQueens(int n) {
        char[][] board = new char[n][n];
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                board[i][j] = '.';
        List<List<String>> res = new ArrayList<List<String>>();
        dfs(board, 0, res);
        return res;
    }
    
    private void dfs(char[][] board, int colIndex, List<List<String>> res) {
        if(colIndex == board.length) {
            res.add(construct(board));
            return;
        }
        
        for(int i = 0; i < board.length; i++) {
            if(validate(board, i, colIndex)) {
                board[i][colIndex] = 'Q';
                dfs(board, colIndex + 1, res);
                board[i][colIndex] = '.';
            }
        }
    }
    
    private boolean validate(char[][] board, int x, int y) {
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < y; j++) {
                if(board[i][j] == 'Q' && (x + j == y + i || x + y == i + j || x == i))
                    return false;
            }
        }
        
        return true;
    }
    
    private List<String> construct(char[][] board) {
        List<String> res = new LinkedList<String>();
        for(int i = 0; i < board.length; i++) {
            String s = new String(board[i]);
            res.add(s);
        }
        return res;
    }
}　　

Python:

class Solution(object):
    def solveNQueens(self, n):
        """
        :type n: int
        :rtype: List[List[str]]
        """
        def dfs(curr, cols, main_diag, anti_diag, result):
            row, n = len(curr), len(cols)
            if row == n:
                result.append(map(lambda x: '.'*x + "Q" + '.'*(n-x-1), curr))
                return
            for i in xrange(n):
                if cols[i] or main_diag[row+i] or anti_diag[row-i+n]:
                    continue
                cols[i] = main_diag[row+i] = anti_diag[row-i+n] = True
                curr.append(i)
                dfs(curr, cols, main_diag, anti_diag, result)
                curr.pop()
                cols[i] = main_diag[row+i] = anti_diag[row-i+n] = False

        result = []
        cols, main_diag, anti_diag = [False]*n, [False]*(2*n), [False]*(2*n)
        dfs([], cols, main_diag, anti_diag, result)
        return result

Python:

# For any point (x,y), if we want the new point (p,q) don't share the same row, column, or diagonal.
# then there must have ```p+q != x+y``` and ```p-q!= x-y```
# the former focus on eliminate 'left bottom right top' diagonal;
# the latter focus on eliminate 'left top right bottom' diagonal

# - col_per_row: the list of column index per row
# - cur_row：current row we are seraching for valid column
# - xy_diff：the list of x-y
# - xy_sum：the list of x+y
class Solution2(object):
    def solveNQueens(self, n):
        """
        :type n: int
        :rtype: List[List[str]]
        """
        def dfs(col_per_row, xy_diff, xy_sum):
            cur_row = len(col_per_row)
            if cur_row == n:
                ress.append(col_per_row)
            for col in range(n):
                if col not in col_per_row and cur_row-col not in xy_diff and cur_row+col not in xy_sum:
                    dfs(col_per_row+[col], xy_diff+[cur_row-col], xy_sum+[cur_row+col])
        ress = []
        dfs([], [], [])
        return [['.'*i + 'Q' + '.'*(n-i-1) for i in res] for res in ress]　　

C++:

class Solution {
public:
    vector<vector<string> > solveNQueens(int n) {
        vector<vector<string> > res;
        vector<int> pos(n, -1);
        solveNQueensDFS(pos, 0, res);
        return res;
    }
    void solveNQueensDFS(vector<int> &pos, int row, vector<vector<string> > &res) {
        int n = pos.size();
        if (row == n) {
            vector<string> out(n, string(n, '.'));
            for (int i = 0; i < n; ++i) {
                out[i][pos[i]] = 'Q';
            }
            res.push_back(out);
        } else {
            for (int col = 0; col < n; ++col) {
                if (isValid(pos, row ,col)) {
                    pos[row] = col;
                    solveNQueensDFS(pos, row + 1, res);
                    pos[row] = -1;
                }
            }
        }
    }
    bool isValid(vector<int> &pos, int row, int col) {
        for (int i = 0; i < row; ++i) {
            if (col == pos[i] || abs(row - i) == abs(col - pos[i])) {
                return false;
            }
        }
        return true;
    }
};

　　

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