[LeetCode] 52. N-Queens II N皇后问题 II

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return the number of distinct solutions to the n-queens puzzle.

Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

51. N-Queens N 的变形，这道题只需要给出不同解法的数量，比51题要简单一些。

解法：回溯Backtracking

Java:

/**
 * don't need to actually place the queen,
 * instead, for each row, try to place without violation on
 * col/ diagonal1/ diagnol2.
 * trick: to detect whether 2 positions sit on the same diagnol:
 * if delta(col, row) equals, same diagnol1;
 * if sum(col, row) equals, same diagnal2.
 */
private final Set<Integer> occupiedCols = new HashSet<Integer>();
private final Set<Integer> occupiedDiag1s = new HashSet<Integer>();
private final Set<Integer> occupiedDiag2s = new HashSet<Integer>();
public int totalNQueens(int n) {
    return totalNQueensHelper(0, 0, n);
}

private int totalNQueensHelper(int row, int count, int n) {
    for (int col = 0; col < n; col++) {
        if (occupiedCols.contains(col))
            continue;
        int diag1 = row - col;
        if (occupiedDiag1s.contains(diag1))
            continue;
        int diag2 = row + col;
        if (occupiedDiag2s.contains(diag2))
            continue;
        // we can now place a queen here
        if (row == n-1)
            count++;
        else {
            occupiedCols.add(col);
            occupiedDiag1s.add(diag1);
            occupiedDiag2s.add(diag2);
            count = totalNQueensHelper(row+1, count, n);
            // recover
            occupiedCols.remove(col);
            occupiedDiag1s.remove(diag1);
            occupiedDiag2s.remove(diag2);
        }
    }

    return count;
}　

Python:

# quick solution for checking if it is diagonally legal
class Solution:
    # @return an integer
    def totalNQueens(self, n):
        self.cols = [False] * n
        self.main_diag = [False] * (2 * n)
        self.anti_diag = [False] * (2 * n)
        return self.totalNQueensRecu([], 0, n)

    def totalNQueensRecu(self, solution, row, n):
        if row == n:
            return 1
        result = 0
        for i in xrange(n):
            if not self.cols[i] and not self.main_diag[row + i] and not self.anti_diag[row - i + n]:
                self.cols[i] = self.main_diag[row + i] = self.anti_diag[row - i + n] = True
                result += self.totalNQueensRecu(solution + [i], row + 1, n)
                self.cols[i] = self.main_diag[row + i] = self.anti_diag[row - i + n] = False
        return result

Python:

# slower solution
class Solution:
    # @return an integer
    def totalNQueens(self, n):
        return self.totalNQueensRecu([], 0, n)

    def totalNQueensRecu(self, solution, row, n):
        if row == n:
            return 1
        result = 0
        for i in xrange(n):
            if i not in solution and reduce(lambda acc, j: abs(row - j) != abs(i - solution[j]) and acc, xrange(len(solution)), True):
                result += self.totalNQueensRecu(solution + [i], row + 1, n)
        return result　

C++:

class Solution {
public:
    int totalNQueens(int n) {
        int res = 0;
        vector<int> pos(n, -1);
        totalNQueensDFS(pos, 0, res);
        return res;
    }
    void totalNQueensDFS(vector<int> &pos, int row, int &res) {
        int n = pos.size();
        if (row == n) ++res;
        else {
            for (int col = 0; col < n; ++col) {
                if (isValid(pos, row, col)) {
                    pos[row] = col;
                    totalNQueensDFS(pos, row + 1, res);
                    pos[row] = -1;
                }
            }
        }
    }
    bool isValid(vector<int> &pos, int row, int col) {
        for (int i = 0; i < row; ++i) {
            if (col == pos[i] || abs(row - i) == abs(col - pos[i])) {
                return false;
            }
        }
        return true;
    }
};

　　

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