51. N-Queens
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.
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
Rust Solution
struct Solution;
impl Solution {
fn solve_n_queens(n: i32) -> Vec<Vec<String>> {
let n = n as usize;
let mut queens: Vec<u32> = vec![];
let mut res = vec![];
let mut column: u32 = 0;
let mut diagonal1: u32 = 0;
let mut diagonal2: u32 = 0;
Self::dfs(
0,
&mut queens,
&mut column,
&mut diagonal1,
&mut diagonal2,
&mut res,
n,
);
res
}
fn dfs(
i: usize,
queens: &mut Vec<u32>,
column: &mut u32,
diagonal1: &mut u32,
diagonal2: &mut u32,
all: &mut Vec<Vec<String>>,
n: usize,
) {
if i == n {
let solution = queens
.iter()
.map(|row| {
(0..n)
.map(|i| if row & (1 << i) > 0 { 'Q' } else { '.' })
.collect::<String>()
})
.collect();
all.push(solution);
} else {
for j in 0..n {
let column_bit = 1 << j;
let diagonal1_bit = 1 << (i + j);
let diagonal2_bit = 1 << (n + i - j);
if column_bit & *column == 0
&& diagonal1_bit & *diagonal1 == 0
&& diagonal2_bit & *diagonal2 == 0
{
*column |= column_bit;
*diagonal1 |= diagonal1_bit;
*diagonal2 |= diagonal2_bit;
queens.push(column_bit);
Self::dfs(i + 1, queens, column, diagonal1, diagonal2, all, n);
queens.pop();
*column &= !column_bit;
*diagonal1 &= !diagonal1_bit;
*diagonal2 &= !diagonal2_bit;
}
}
}
}
}
#[test]
fn test() {
let n = 4;
let res = vec_vec_string![
[".Q..", "...Q", "Q...", "..Q."],
["..Q.", "Q...", "...Q", ".Q.."]
];
assert_eq!(Solution::solve_n_queens(n), res);
}
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