[leetcode-51-N-Queens]

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.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

In this problem, we can go row by row, and in each position, we need to check if the column, the 45° diagonal and the 135° diagonal had a queen before.

Solution A: Directly check the validity of each position, 12ms:

class Solution {
public:
    std::vector<std::vector<std::string> > solveNQueens(int n) {
        std::vector<std::vector<std::string> > res;
        std::vector<std::string> nQueens(n, std::string(n, '.'));
        solveNQueens(res, nQueens, 0, n);
        return res;
    }
private:
    void solveNQueens(std::vector<std::vector<std::string> > &res, std::vector<std::string> &nQueens, int row, int &n) {
        if (row == n) {
            res.push_back(nQueens);
            return;
        }
        for (int col = 0; col != n; ++col)
            if (isValid(nQueens, row, col, n)) {
                nQueens[row][col] = 'Q';
                solveNQueens(res, nQueens, row + 1, n);
                nQueens[row][col] = '.';
            }
    }
    bool isValid(std::vector<std::string> &nQueens, int row, int col, int &n) {
        //check if the column had a queen before.
        for (int i = 0; i != row; ++i)
            if (nQueens[i][col] == 'Q')
                return false;
        //check if the 45° diagonal had a queen before.
        for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; --i, --j)
            if (nQueens[i][j] == 'Q')
                return false;
        //check if the 135° diagonal had a queen before.
        for (int i = row - 1, j = col + 1; i >= 0 && j < n; --i, ++j)
            if (nQueens[i][j] == 'Q')
                return false;
        return true;
    }
};

参考：

https://discuss.leetcode.com/topic/13617/accepted-4ms-c-solution-use-backtracking-and-bitmask-easy-understand