[LeetCode] 52. N-Queens II

思路:
维护三个数组, 分别记录同列, 左上对角线, 右上对角线是否有Queen, 对角线数组要2n长, 这样就能装下所有col + row的可能, 比较方便. 主要算法依然是DFS, 很标准.

void dfs(int& res, int row, int& n, vector<bool>& colCheck,
            vector<bool>& leftDiagCheck, vector<bool>& rightDiagCheck) {
    if (row == n) {
        res++;
        return;
    }
    for (int col = 0; col < n; col++) {
        int leftDiag = col - row + n;
        int rightDiag = col + row;
        if (colCheck[col] || leftDiagCheck[leftDiag] || rightDiagCheck[rightDiag])
            continue;
        colCheck[col] = leftDiagCheck[leftDiag] = rightDiagCheck[rightDiag] = true;
        dfs(res, row + 1, n, colCheck, leftDiagCheck, rightDiagCheck);
        colCheck[col] = leftDiagCheck[leftDiag] = rightDiagCheck[rightDiag] = false;
    }
}

int totalNQueens(int n) {
    int res = 0;
    vector<bool> colCheck(n);
    vector<bool> leftDiagCheck(2 * n);
    vector<bool> rightDiagCheck(2 * n);
    dfs(res, 0, n, colCheck, leftDiagCheck, rightDiagCheck);
    return res;
}