`

N-Queens II

阅读更多
Follow up for N-Queens problem.

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


同样是N皇后的问题,要求我们输出有几种结果,思路同N-Queens是一样的,这里就不在赘述了,代码如下:
public class Solution {
    private int result = 0;
    public int totalNQueens(int n) {
        int[] colInRow = new int[n];
        getTotalNQueens(0, n, colInRow);
        return result;
    }
    
    private void getTotalNQueens(int row, int n, int[] colInRow) {
        if(row == n) {
            result ++;
        } else {
            for(int i = 0; i < n; i++) {
                colInRow[row] = i;
                if(isValid(row, colInRow)) {
                    getTotalNQueens(row + 1, n, colInRow);
                }
            }
        }
    }
    
    private boolean isValid(int row, int[] colInRow) {
        for(int i = 0; i < row; i++) {
            if(colInRow[i] == colInRow[row] || Math.abs(colInRow[i] - colInRow[row]) == row - i)
                return false;
        }
        return true;
    }
}
分享到:
评论

相关推荐

Global site tag (gtag.js) - Google Analytics