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LeetCode "N-Queens II"

Classic recursion\pruning 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;    }};
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