[leetcode] N-Queens II

最新推荐文章于 2022-06-06 10:35:33 发布

原创最新推荐文章于 2022-06-06 10:35:33 发布 · 278 阅读

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本文深入探讨了N皇后问题的后续挑战，即计算所有独特的解决方案总数。通过介绍两种方法来解决这个问题，文章不仅提供了解决方案的数量，还解释了如何在不显示棋盘配置的情况下实现这一目标。

from : https://leetcode.com/problems/n-queens-ii/

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

public class Solution {
    private int n;
    private int total;
    private int[] queue;
    
    public int totalNQueens(int n) {
        init(n);
        
        traverse(0);
        
        return total;
    }
    
    private void traverse(int k) {
        if(k == n) {
            ++total;
            return;
        }
        
        for(int i=0; i<n; ++i) {
            boolean valid = true;
            for(int j=0; j<k; ++j) {
                if(queue[j] == i || Math.abs(queue[j] - i) == k - j) {
                    valid = false;
                    break;
                }
            }
            
            if(valid) {
                queue[k] = i;
                traverse(k+1);
            }
        }
        
    }
    
    private void init(int n) {
        total = 0;
        queue = new int[n];
        this.n = n;
    }
}

public class Solution {

	private int N, ans;
	private int[] position;

	public int totalNQueens(int n) {
		if (n < 1) {
			return 0;
		}
		init(n);
		takeRow(0);
		return ans;
	}

	private void takeRow(int k) {
		if (k == N) {
			++ans;
			return;
		}
		for (int j = 0; j < N; ++j) {
			boolean valid = true;
			for (int i = 0; i < k && valid; ++i) {
				if (position[i] == j || Math.abs(position[i] - j) == k - i) {
					valid = false;
				}
			}
			if (valid) {
				position[k] = j;
				takeRow(k + 1);
			}
		}
	}

	private void init(int n) {
		ans = 0;
		N = n;
		position = new int[n];
	}
}