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LeetCode "N-Queens II"

2014-07-21 06:49 351 查看
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
;

go(A, 0, n);

delete[] A;
return ret;
}
};


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