51. N-Queens

Posted on Edited on In

51. N-Queens

Description

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:

1
2
3
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:

1
2
Input: n = 1
Output: [["Q"]]

Constraints:

  • 1 <= n <= 9

Hints/Notes

Solution

Language: C++

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
class Solution {
public:
    vector<int> rows, cols, diag1, diag2;
    vector<vector<string>> res;
    int n;

    vector<vector<string>> solveNQueens(int n) {
        this->n = n;
        rows.resize(n, 0); cols.resize(n, 0); diag1.resize(2 * n, 0); diag2.resize(2 * n - 1, 0);
        dfs(0);
        return res;
    }

    void dfs(int index) {
        if (index == n) {
            vector<string> ans;
            for (int i = 0; i < n; i++) {
                string cow = string(rows[i], '.') + 'Q' + string(n - 1 - rows[i], '.');
                ans.push_back(cow);
            }
            res.push_back(ans);
            return;
        }
        for (int i = 0; i < n; i++) {
            if (!cols[i] && !diag1[i - index + n] && !diag2[index + i]) {
                rows[index] = i;
                cols[i] = diag1[i - index + n] = diag2[index + i] = true;
                dfs(index + 1);
                cols[i] = diag1[i - index + n] = diag2[index + i] = false;
            }
        }
    }
};