Maximum Product of all but one Entries

Solution from EPI book
O(N) time and O(N) space  
public static int findBiggestProductNMinusOneProduct(List<Integer> A) {
    // Build forward product L, and backward product R.
    List<Integer> L = new ArrayList<>(A.size());
    List<Integer> R = new ArrayList<>(A.size());
    fill(L, A.size(), 0);
    fill(R, A.size(), 0);
    partialSum(A.iterator(), L.listIterator(), BinaryOperators.MULTIPLIES);
    partialSum(revListIterator(A), revListIterator(R),
        BinaryOperators.MULTIPLIES);

    // Find the biggest product of (n - 1) numbers.
    int maxProduct = Integer.MIN_VALUE;
    for (int i = 0; i < A.size(); ++i) {
      int forward = i > 0 ? L.get(i - 1) : 1;
      int backward = i + 1 < A.size() ? R.get(i + 1) : 1;
      maxProduct = max(maxProduct, forward * backward);
    }
    return maxProduct;

  }
O(N) time and O(1) space
  public static int findBiggestNMinusOneProduct(int[] A) {
    int zeroCount = 0, posCount = 0, negCount = 0;
    int zeroIdx = -1, sNegIdx = -1, bNegIdx = -1, sPosIdx = -1;

    for (int i = 0; i < A.length; ++i) {
      if (A[i] < 0) {
        ++negCount;
        if (sNegIdx == -1 || A[i] < A[sNegIdx]) {
          sNegIdx = i;
        }
        if (bNegIdx == -1 || A[bNegIdx] < A[i]) {
          bNegIdx = i;
        }
      } else if (A[i] == 0) {
        zeroIdx = i;
        ++zeroCount;
      } else { // A[i] > 0.
        ++posCount;
        if (sPosIdx == -1 || A[i] < A[sPosIdx]) {
          sPosIdx = i;
        }
      }
    }

    // Try to find a number whose elimination could maximize the product of
    // the remaining (n - 1) numbers.
    int x; // Stores the idx of eliminated one.
    if (zeroCount >= 2) {
      return 0;
    } else if (zeroCount == 1) {
      if ((negCount & 1) > 0) {
        return 0;
      } else {
        x = zeroIdx;
      }
    } else {
      if ((negCount & 1) > 0) { // Odd number negative.
        x = bNegIdx;
      } else { // Even number negative.
        if (posCount > 0) {
          x = sPosIdx;
        } else {
          x = sNegIdx;
        }
      }
    }

    int product = 1;
    for (int i = 0; i < A.length; ++i) {
      if (i != x) {
        product *= A[i];
      }
    }
    return product;

  }