N Queens

LeetCode #51

Description:

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:

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

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

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

Idea:

N Queens与N Queens II很类似，II需要把结果存下来，而不是count增加。

Mark的是occupy哪列和哪些对角线，Backtrack也是这些。II的时候，还要存暂时的结果，不过不需要backtrack temporary result.

Code:

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<bool> column_occupy(n, false);
        vector<bool> diag_occupy(2*n-1, false);
        vector<bool> anti_diag_occupy(2*n-1, false);
        vector<vector<string>> result;
        vector<int> queens(n, -1);
        DFS(0, queens, result, column_occupy, diag_occupy, anti_diag_occupy);
        return result;
    }

    void DFS(int row, vector<int>& queens, vector<vector<string>>& result, vector<bool>& column_o, vector<bool>& diag_o, vector<bool>& anti_diag_o){
        int N=column_o.size();
        if(row==N){
            vector<string> solution(N, string(N, '.'));
            for(int i=0; i<row; ++i){
                solution[i][queens[i]]='Q';
            }
            result.push_back(solution);
            return;
        }

        // j is column index
        for(int j=0; j<N; ++j){
            if(!column_o[j] && !diag_o[row+j] && !anti_diag_o[N-1-row+j]){
                queens[row]=j;
                column_o[j]=diag_o[row+j]=anti_diag_o[N-1-row+j]=true;
                DFS(row+1, queens, result, column_o, diag_o, anti_diag_o);
                column_o[j]=diag_o[row+j]=anti_diag_o[N-1-row+j]=false;
            }
        }
    }
};