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面试题 08.12. 八皇后

English Version

题目描述

设计一种算法，打印 N 皇后在 N × N 棋盘上的各种摆法，其中每个皇后都不同行、不同列，也不在对角线上。这里的“对角线”指的是所有的对角线，不只是平分整个棋盘的那两条对角线。

注意：本题相对原题做了扩展

示例:

 输入：4
 输出：[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
 解释: 4 皇后问题存在如下两个不同的解法。
[
 [".Q..",  // 解法 1
  "...Q",
  "Q...",
  "..Q."],

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

解法

深度优先搜索 + 剪枝。

Python3

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        res = []
        g = [['.'] * n for _ in range(n)]
        col = [False] * n
        dg = [False] * (2 * n)
        udg = [False] * (2 * n)

        def dfs(u):
            if u == n:
                res.append([''.join(item) for item in g])
                return
            for i in range(n):
                if not col[i] and not dg[u + i] and not udg[n - u + i]:
                    g[u][i] = 'Q'
                    col[i] = dg[u + i] = udg[n - u + i] = True
                    dfs(u + 1)
                    g[u][i] = '.'
                    col[i] = dg[u + i] = udg[n - u + i] = False

        dfs(0)
        return res

Java

class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> res = new ArrayList<>();
        String[][] g = new String[n][n];
        for (int i = 0; i < n; ++i) {
            String[] t = new String[n];
            Arrays.fill(t, ".");
            g[i] = t;
        }
        // 列是否已经有值
        boolean[] col = new boolean[n];
        // 斜线是否已经有值
        boolean[] dg = new boolean[2 * n];
        // 反斜线是否已经有值
        boolean[] udg = new boolean[2 * n];
        // 从第一行开始搜索
        dfs(0, n, col, dg, udg, g, res);
        return res;
    }

    private void dfs(int u, int n, boolean[] col, boolean[] dg, boolean[] udg, String[][] g,
        List<List<String>> res) {
        if (u == n) {
            List<String> t = new ArrayList<>();
            for (String[] e : g) {
                t.add(String.join("", e));
            }
            res.add(t);
            return;
        }
        for (int i = 0; i < n; ++i) {
            if (!col[i] && !dg[u + i] && !udg[n - u + i]) {
                g[u][i] = "Q";
                col[i] = dg[u + i] = udg[n - u + i] = true;
                dfs(u + 1, n, col, dg, udg, g, res);
                g[u][i] = ".";
                col[i] = dg[u + i] = udg[n - u + i] = false;
            }
        }
    }
}

C++

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<vector<string>> res;
        vector<string> g(n, string(n, '.'));
        vector<bool> col(n, false);
        vector<bool> dg(2 * n, false);
        vector<bool> udg(2 * n, false);
        dfs(0, n, col, dg, udg, g, res);
        return res;
    }

    void dfs(int u, int n, vector<bool>& col, vector<bool>& dg, vector<bool>& udg, vector<string>& g, vector<vector<string>>& res) {
        if (u == n) {
            res.push_back(g);
            return;
        }
        for (int i = 0; i < n; ++i) {
            if (!col[i] && !dg[u + i] && !udg[n - u + i]) {
                g[u][i] = 'Q';
                col[i] = dg[u + i] = udg[n - u + i] = true;
                dfs(u + 1, n, col, dg, udg, g, res);
                g[u][i] = '.';
                col[i] = dg[u + i] = udg[n - u + i] = false;
            }
        }
    }
};

Go

func solveNQueens(n int) [][]string {
	res := [][]string{}
	g := make([][]string, n)
	for i := range g {
		g[i] = make([]string, n)
		for j := range g[i] {
			g[i][j] = "."
		}
	}
	col := make([]bool, n)
	dg := make([]bool, 2*n)
	udg := make([]bool, 2*n)
	dfs(0, n, col, dg, udg, g, &res)
	return res
}

func dfs(u, n int, col, dg, udg []bool, g [][]string, res *[][]string) {
	if u == n {
		t := make([]string, n)
		for i := 0; i < n; i++ {
			t[i] = strings.Join(g[i], "")
		}
		*res = append(*res, t)
		return
	}
	for i := 0; i < n; i++ {
		if !col[i] && !dg[u+i] && !udg[n-u+i] {
			g[u][i] = "Q"
			col[i], dg[u+i], udg[n-u+i] = true, true, true
			dfs(u+1, n, col, dg, udg, g, res)
			g[u][i] = "."
			col[i], dg[u+i], udg[n-u+i] = false, false, false
		}
	}
}

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