LeetCode: N-Queens II

Tingmei

于 2012-10-09 00:04:48 发布

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分类专栏：算法 C++

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本文介绍了一种递归算法解决N皇后问题，并计算所有可能的解决方案数量。通过检查每一行皇后的位置来确保没有冲突，该算法有效地计算了不同布局的数量。

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Follow up for N-Queens problem.

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

递归解法：

class Solution {
public:
    bool check(int row, int* place)
    {
        for (int i = 0; i < row; ++i)
        {
            int diff = abs(place[i] - place[row]);
            if (diff == 0 || diff == row - i)
                return false;
        }
        return true;
    }
    
    void placeQueens(int row, int n, int &count, int* place)
    {
        if (row == n)
        {
            ++count;
            return;
        }
        
        for (int i = 0; i < n; ++i)
        {
            place[row] = i;
            if (check(row, place))
                placeQueens(row+1, n, count, place);
        }
    }
    
    int totalNQueens(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int* place = new int[n];
        int count = 0;
        placeQueens(0, n, count, place);
        return count;
    }
};