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:
Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
public List<List<String>> solveNQueens(int n) {
List<List<String>> result = new ArrayList<>();
HashSet<Integer> cols = new HashSet<>();
StringBuilder sb = new StringBuilder();
LinkedList<int[]> qs = new LinkedList<>();
for(int i = 0; i < n * n; i++) sb.append('.');
backtrack(result, cols, qs, sb, n, 0);
return result;
}
public void backtrack(List<List<String>> result, HashSet<Integer> cols, LinkedList<int[]> qs, StringBuilder sb, int n, int row) {
out:for(int i = 0; i < n; i++) {
if(cols.contains(i)) continue;
int sum = row + i, diff = row - i;
for(int[] q : qs) {
if(q[0] + q[1] == sum || q[0] - q[1] == diff)
continue out;
}
cols.add(i);
qs.add(new int[] {row, i});
sb.setCharAt(row*n+i, 'Q');
if(row == n-1) {
ArrayList<String> temp = new ArrayList<>();
for(int j = 0; j < n; j++) {
temp.add(sb.substring(j*n, j*n+n));
}
result.add(temp);
}
backtrack(result, cols, qs, sb, n, row+1);
qs.removeLast();
sb.setCharAt(row*n+i, '.');
cols.remove(i);
}
