leetcode 127: N-Queens II

最新推荐文章于 2022-06-06 22:03:40 发布

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最新推荐文章于 2022-06-06 22:03:40 发布

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本文探讨了N皇后问题的变种，旨在计算而非输出棋盘配置，并提出了一种优化方案：使用一维数组替代二维矩阵来节省时间和空间。通过递归方法实现了对所有可能解的计数。

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N-Queens IIMar 20 '12

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions. A good optimization example: when the request result is simplified, we may use reduced data struct to save time and space.

won't pass the big dataset until use one dimension array to replace the matrix in queen I.

class Solution {
public:
    int totalNQueens(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> board(n, -1);
        int sum = 0;
        queenRec(board, 0, sum);
        return sum;
    }
private:
    void queenRec(vector<int>& board, int level, int & sum) {
        int n = board.size();
        if(level==n) {
            ++sum;
            return;
        }
        
        for(int i=0; i<n; i++) {
            if(check(board, level, i)) {
                board[level] = i;
                queenRec(board, level+1, sum);
                board[level] = -1;
            }
        }
    }
    
    bool check(const vector<int>& cur, int row, int col) {
        int n= cur.size();
        for(int i=0; i<row; i++) {
            if(cur[i]==col || row-i == abs(col-cur[i]) ) return false; 
        }
        return true;
    }
};