For example, given the array [2,3,-2,4], the contiguous subarray [2,3] has the largest product = 6.

Solution: One Pass O(n) time.

So for each index i, max(i) = the maximum product ending at i, so we have

max(i) = max(max(i-1)*nums[i], nums[i]) //nums[i]>=0

since you can either append from the previous one, or just start a new contiguous subarray.

Be careful on negative nums, if nums[i] < 0, the equation now becomes:

max(i) = max(min(i-1)*nums[i], nums[i]) //nums[i]<0

public int maxProduct(int[] nums) {
    if (nums == null || nums.length == 0) {
        return 0;
    }
    int maxProduct = Integer.MIN_VALUE;
    int preMax = 1;
    int preMin = 1;
    for (int i = 0; i < nums.length; i++) {
        int max, min;
        if (nums[i] >= 0) {
            max = Math.max(preMax * nums[i], nums[i]);
            min = Math.min(preMin * nums[i], nums[i]);
        } else {
            max = Math.max(preMin * nums[i], nums[i]);
            min = Math.min(preMax * nums[i], nums[i]);
        }
        maxProduct = Math.max(maxProduct, max);
        preMax = max;
        preMin = min;
    }
    return maxProduct;
}