LeetCode "N-Queens II"

Classic recursionpruning problem. We can use O(n) space: A[i] = j means [i,j] is occupied.

class Solution {
public:
    int ret;
    bool isValid(int *A, int r)
    {
        for (int i = 0; i < r; i++)
            if ((abs(A[i] - A[r]) == abs(i - r) ||    A[i] == A[r]))
                return false;
        return true;
    }

    void go(int *A, int r, int n)
    {
        if (r == n)
        {
            ret++;
            return;
        }
        for (int i = 0; i < n; i++)
        {
            A[r] = i;
            if (isValid(A, r))    go(A, r + 1, n);
        }
    }
    int totalNQueens(int n) {
        ret = 0;
        int *A = new int[n];

        go(A, 0, n);

        delete[] A;
        return ret;
    }
};