应用Kadane算法，以获得最大的产品子阵似乎棘手。虽然我能够获得最大的产品，我不能得到正确的范围最大的产品子数组。

Applying Kadane algorithm to get max product subarray seems tricky. While I am able to get the max product, I am not really getting the correct range of the max product subarray.

http://www.geeksforgeeks.org/maximum-product-subarray/解释的方式来获得最大的产品，但我不明白，我们怎么能得到子数组的范围。

http://www.geeksforgeeks.org/maximum-product-subarray/ explains the way to get the max product but I dont get how we can get the range of the subarray.

有人可以帮助我了解的范围内的问题？这是一个标准的面试问题，我想确保我的理解对产品的情况下，而不是只是说，最大总和子数组可以修改回答最多的产品子阵情况下的逻辑。

Can someone help me understand the range issue? This is a standard interview question and I want to make sure I understand the logic for the product case instead of just saying that the max sum subarray can be modified to answer max product subarray case.

谢谢！

推荐答案

它基本上有三种情况：

您需要有两个变量：

分占据着最小值至今

最高持有的最大值到现在。

max which holds the maximum value till now.

现在的情况下3 分和最高将被复位到1。

Now for case 3 min and max will be reset to 1.

有关案例1 ：最高将最大* A [1] 和分将是最小的分钟* A [1] 和 1。

For case 1: max will be max * a[i] and min will be minimum of min*a[i] and 1.

有关案例2 ：最高将是最大的的[I] *分钟和 1 ，而分值将是最大* A [1]。

For case 2: max will be maximum of a[i] * min and 1, but the min value will be max * a[i].

下面是code：

private static int getMaxProduct(int[] a){
    int minCurrent = 1, maxCurrent = 1, max = Integer.MIN_VALUE;
    for (int current : a) {
        if (current > 0) {
            maxCurrent = maxCurrent * current;
            minCurrent = Math.min(minCurrent * current, 1);
        } else if (current == 0) {
            maxCurrent = 1;
            minCurrent = 1;
        } else {
            int x = maxCurrent;
            maxCurrent = Math.max(minCurrent * current, 1);
            minCurrent = x * current;
        }
        if (max < maxCurrent) {
            max = maxCurrent;
        }
    }
    //System.out.println(minCurrent);
    return max;
}