Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
[img]http://www.leetcode.com/wp-content/uploads/2012/03/8-queens.png[/img]
同样是N皇后的问题,要求我们输出有几种结果,思路同[url=http://kickcode.iteye.com/blog/2280861]N-Queens[/url]是一样的,这里就不在赘述了,代码如下:
Now, instead outputting board configurations, return the total number of distinct solutions.
[img]http://www.leetcode.com/wp-content/uploads/2012/03/8-queens.png[/img]
同样是N皇后的问题,要求我们输出有几种结果,思路同[url=http://kickcode.iteye.com/blog/2280861]N-Queens[/url]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;
}
}
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