[Leetcode 91] 51 N-Queens

Problem:

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.."]
]

Analysis:

Classic DFS problem. During the enumeration, we set the '.' to 'Q' level by level. So we don't need to judge whether there is level collision. We use a one-dimension array to keep track of whether a column has already contains a Queen. And for diagonal and anti-diagonal, we make use of the property that slope of diagonal elements is 1 and of anti-diagonal is -1 to find those positions to be checked and then check. 

There are still many optimizations can be done to further improve the solution

Code:

class Solution {
public:
    int dim, *col;
    vector<vector<string> > res;

    vector<vector<string> > solveNQueens(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        dim = n;
        res.clear();
        
        string s(n, '.');
        col = new int[n];
        vector<string> tmp;
        for (int i=0; i<n; i++) {
            col[i] = 0;
            tmp.push_back(s);
        }

        dfs(tmp, 0);
        
        delete [] col;
        
        return res;
    }
    
    
private:
    void dfs(vector<string> &s, int r) {
        if (r == dim) {
            res.push_back(s);
            return ;
        }

        for (int i=0; i<dim; i++) {
            if (col[i] == 0 && isValid(s, r, i)) {
                col[i] = 1;
                s[r][i] = 'Q';
                dfs(s, r+1);
                s[r][i] = '.';
                col[i] = 0;
            }
        }
    }


    bool isValid(vector<string> &s, int r, int c) {
        for (int i=0; i<dim; i++) {
            for (int j=0; j<dim; j++) {
                if ((abs(i-r) == abs(j-c)) || abs(i-r) == -abs(j-c)) {
                    if (s[i][j] == 'Q')
                        return false;
                }
            }
        }

        return true;
    }
};