191. Maximum Product Subarray

这道题是Maximum Subarray的follow up
应为乘法所以要考虑负负得正
所以同时记录最小值与最大值

max_products[i] = Math.max(Math.max(max_products[i-1] * nums[i],min_products[i-1]*nums[i]), nums[i])；
min_products[i] = Math.min(Math.min(max_products[i-1] * nums[i],min_products[i-1]*nums[i]), nums[i]);

public class Solution {
    /**
     * @param nums: An array of integers
     * @return: An integer
     */
    public int maxProduct(int[] nums) {
        // write your code here
        int len = nums.length;
        Integer[] max_products = new Integer[len];
        Integer[] min_products = new Integer[len];
       
        max_products[0] = nums[0];
        min_products[0] = nums[0];
        int max = max_products[0];
        for(int i = 1; i < len; i++) {
            max_products[i] = Math.max(Math.max(max_products[i-1] * nums[i],min_products[i-1]*nums[i]), nums[i]);
            min_products[i] = Math.min(Math.min(max_products[i-1] * nums[i],min_products[i-1]*nums[i]), nums[i]);
            max = Math.max(max_products[i], max);
        }
       
        return max;
    }
}