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Leetcode dfs N-Queens

2024-07-21 20:58:07   220人阅读

N-Queens

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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.

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

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

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


题意：返回n皇后问题的所有可能解
思路：dfs
要求给出所有可能解的题，一般采用dfs 暴力枚举
这道题设置了额外的三个bool数组分别表示某一列，
某一主对角线，某一副对角线是否已经有皇后占据着。
由某个坐标(i,j)，可得到它的主对角线是i+j，它的副对角线是i-j，
由于i-j可能是负数，所以我们给它加上 n,保证它不为负数
复杂度：O(n!),空间O(n)

vector<vector<string> >res;
int _n;
vector<bool> is_col_chosen;
vector<bool> is_main_diag_chosen;
vector<bool> is_anti_diag_chosen;


bool is_chosen(int i, int j){
	return is_col_chosen[j] || is_main_diag_chosen[i + j] || is_anti_diag_chosen[i - j + _n];
}


void dfs(int nth_queen, vector<int> &columns){
	if(nth_queen == _n){
		vector<string> one_res(_n, string(_n, '.'));
		int row = 0;
		for_each(one_res.begin(), one_res.end(), [&columns, &row](string &one_row){
			one_row[columns[row++]] = 'Q';
		});
		res.push_back(one_res);
	}
	for(int j = 0; j < _n; ++j){
		if(!is_chosen(nth_queen, j)){
			is_col_chosen[j] = is_main_diag_chosen[nth_queen + j] = is_anti_diag_chosen[nth_queen - j + _n] = true;
			columns.push_back(j);
			dfs(nth_queen + 1,columns);
			columns.pop_back();
			is_col_chosen[j] = is_main_diag_chosen[nth_queen + j] = is_anti_diag_chosen[nth_queen - j + _n] = false;
		}
	}
}


vector<vector<string> > solveNQueens(int n){
	_n = n;
	is_col_chosen = is_main_diag_chosen = is_anti_diag_chosen = vector<bool>(2 * n, false);
	vector<int> columns;
	dfs(0, columns);
	return res;
}


Leetcode dfs N-Queens


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