【Lintcode】033.N-Queens

题目：

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

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

[
  // Solution 1
  [".Q..",
   "...Q",
   "Q...",
   "..Q."
  ],
  // Solution 2
  ["..Q.",
   "Q...",
   "...Q",
   ".Q.."
  ]
]

 

题解：

　　此题需要注意的是对角线因素，不仅不能对角线相邻，而且不能在一条对角线上，中间隔着一个也不行！刚开始没注意，花了一个多小时才明白过来。。。

Solution 1 ()

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

 Solution 1.2 () from here

class Solution {
private:
    vector<vector<string> > res;
public:
    vector<vector<string> > solveNQueens(int n) {
        vector<string>cur(n, string(n,'.'));
        helper(cur, 0);
        return res;
    }
    void helper(vector<string> &cur, int row)
    {
        if(row == cur.size())
        {
            res.push_back(cur);
            return;
        }
        for(int col = 0; col < cur.size(); col++)
            if(isValid(cur, row, col))
            {
                cur[row][col] = 'Q';
                helper(cur, row+1);
                cur[row][col] = '.';
            }
    }
     
    //判断在cur[row][col]位置放一个皇后，是否是合法的状态
    //已经保证了每行一个皇后，只需要判断列是否合法以及对角线是否合法。
    bool isValid(vector<string> &cur, int row, int col)
    {
        //列
        for(int i = 0; i < row; i++)
            if(cur[i][col] == 'Q')return false;
        //右对角线(只需要判断对角线上半部分，因为后面的行还没有开始放置)
        for(int i = row-1, j=col-1; i >= 0 && j >= 0; i--,j--)
            if(cur[i][j] == 'Q')return false;
        //左对角线(只需要判断对角线上半部分，因为后面的行还没有开始放置)
        for(int i = row-1, j=col+1; i >= 0 && j < cur.size(); i--,j++)
            if(cur[i][j] == 'Q')return false;
        return true;
    }
};

Solution 1.3 ()

class Solution2 {
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;
    }
};