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:
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:
Input: n = 1
Output: [["Q"]]
Constraints:
- 1 <= n <= 9
class Solution {
public List<List<String>> solveNQueens(int n) {
char[][] board = new char[n][n];
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
board[i][j] = '.';
List<List<String>> res = new ArrayList<List<String>>();
dfs(board, 0, res);
return res;
}
private void dfs(char[][] board, int colIndex, List<List<String>> res) {
if(colIndex == board.length) {
res.add(construct(board));
return;
}
for(int i = 0; i < board.length; i++) {
if(validate(board, i, colIndex)) {
board[i][colIndex] = 'Q';
dfs(board, colIndex + 1, res);
board[i][colIndex] = '.';
}
}
}
private boolean validate(char[][] board, int x, int y) {
for(int i = 0; i < board.length; i++) {
for(int j = 0; j < y; j++) {
if(board[i][j] == 'Q' && (x + j == y + i || x + y == i + j || x == i))
return false;
}
}
return true;
}
private List<String> construct(char[][] board) {
List<String> res = new LinkedList<String>();
for(int i = 0; i < board.length; i++) {
String s = new String(board[i]);
res.add(s);
}
return res;
}
}