51. N-Queens --- H !!!!

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

Given an integer n, return all distinct solutions to the n-queens puzzle. You may return the answer in any order.

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 1:

Input: n = 4
Output: [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above

Example 2:

Input: n = 1
Output: [["Q"]]

 

Constraints:

• 1 <= n <= 9

A:
 --------------------2 nd pass----------------------
DFS ----------  Use 2D array
出现的问题，还是在于，测试diagonal的时候， 两条diagonal的表达有问题。

并不是对角线上的2条才需要检查。 左右的diagonal方向都需要。

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<vector<char>> board(n, vector<char>(n, '.'));
        vector<vector<string>> res;
        dfs(board, res, 0);
        return res;
    }
private:
    void dfs(vector<vector<char>>& board, vector<vector<string>>& res, int row) {
        int n = board.size();
        if (row == n) {
            vector<string> curRes;
            for (int i = 0; i < n; i++) {
                string tmp(board[i].begin(), board[i].end());
                curRes.push_back(tmp);
            }
            res.push_back(curRes);
        } else {
            for (int j = 0; j < n; j++) {
                board[row][j] = 'Q';
                if (isValid(board, row, j))
                    dfs(board, res, row + 1);
                board[row][j] = '.';
            }
        }
    }
    bool isValid(vector<vector<char>>& board, int row, int col) {
        int n = board.size();
        // check each row i
        for (int j = 0; j < n; j++) {
            if (j != col && board[row][j] == 'Q')
                return false;
        }
        // now check colum : col
        for (int i = 0; i < n; i++) {
            if (i != row && board[i][col] == 'Q')
                return false;
        }
        // check diagonal
        for (int i = 0; i < n; i++) {
            int j = i + (col - row);
            if (j >= 0 && j < n && i != row && board[i][j] == 'Q')
                return false;
        }
        // check reverse diagonal
        for (int i = 0; i < n; i++) {
            int j = (col + row) - i;
            if (j >= 0 && j < n && i != row && board[i][j] == 'Q')
                return false;
        }
        return true;
    }
};

--------------3rd pass---------------------

这里用了2D array，  如果用一维数组表示的话（类似于N-Queen II) ，

方法是：一维数组的下标表示横坐标（哪一行），而数组的值表示纵坐标（哪一列）。

Mistakes: