Question

Formatted question description: https://leetcode.ca/all/52.html

 Follow up for N-Queens problem.

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

Algorithm

In fact, I think the order of the two Queens-question should be reversed. The previous question is a little more complicated than this one. There is no difference in essence between the two. Both have to be solved by backtracking. If you understand the previous question The idea of the question, this question only needs to make a small change, no need to ask for the specific queen’s pendulum, just need to increase the counter by one every time a solution is generated.

Code

Java

public class N_Queens_II {

	// just simply return list.size(), seems costly if only count is needed

	public class Solution {

		List<List<String>> list = new ArrayList<List<String>>();

		public int totalNQueens(int n) {

			if (n <= 0)
				return 0;

			// index is row-number, place[index] is position of queen on that
			// row
			int[] place = new int[n];

			dfs(n, 0, place);

			return list.size();
		}

		public void dfs(int n, int currentRow, int[] place) {
			if (currentRow == n) {
				// save to list
				saveToList(place);
				return;
			}

			for (int i = 0; i < n; i++) {

				if (isValid(i, currentRow, place)) {
					place[currentRow] = i; // 1 meaning place queen
					dfs(n, currentRow + 1, place);
					place[currentRow] = i; // undo place queen
				}
			}
		}

		public boolean isValid(int currentRowPosition, int currentRow,
				int[] place) {
			// check all previous rows
			for (int i = 0; i < currentRow; i++) {
			    if (place[i] == currentRowPosition
						|| currentRow - i == (int) Math.abs(place[i] - currentRowPosition))
					return false;
			}

			return true;
		}

		public void saveToList(int[] place) {

			int len = place.length;
			List<String> one = new ArrayList<String>(len);

			// construct string of n dots...
			String row = "";
			for (int i = 0; i < len; i++) {
				row += ".";
			}

			for (int i = 0; i < len; i++) {
				int rowPosition = place[i]; // at position place[i] on row i
				one.add(row.substring(0, rowPosition) + "Q"
						+ row.substring(rowPosition + 1));
			}

			list.add(one);
		}
	}

}

// OJ: https://leetcode.com/problems/n-queens-ii/
// Time: O(N!)
// Space: O(N)
class Solution {
public:
    int totalNQueens(int n) {
        vector<bool> col(n), hill(2 * n - 1), dale(2 * n - 1);
        function<int(int)> dfs = [&](int i) { // for each i-th row
            if (i == n) return 1;
            int ans = 0;
            for (int j = 0; j < n; ++j) { // try putting on the j-th column
                int h = i + j, d = i + n - 1 - j;
                if (col[j] || hill[h] || dale[d]) continue;
                col[j] = hill[h] = dale[d] = true;
                ans += dfs(i + 1);
                col[j] = hill[h] = dale[d] = false;
            }
            return ans;
        };
        return dfs(0);
    }
};

class Solution(object):
  def totalNQueens(self, n):
    """
    :type n: int
    :rtype: int
    """

    def dfs(path, n):
      if len(path) == n:
        return 1
      res = 0
      for i in range(n):
        if i not in path and isValidQueen(path, i):
          path.append(i)
          res += dfs(path, n)
          path.pop()
      return res

    def isValidQueen(path, k):
      for i in range(len(path)):
        if abs(k - path[i]) == abs(len(path) - i):
          return False
      return True

    return dfs([], n)

52 - N-Queens II

Question

Algorithm

Code

All Problems

All Solutions