LeetCode 152. Maximum Product Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest product.

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

分别创建一个最大、最小数组用于记录数据，当数据项小于零从最小数组中选择元素，当数据项大于零时从最大数组中选择元素

java

class Solution {
    public int maxProduct(int[] nums) {
        if (nums == null || nums.length ==0) {
            return 0;
        }
        int[] max = new int[nums.length];
        int[] min = new int[nums.length];
        max[0] = nums[0]; min[0] = nums[0];
        int val = nums[0];
        for (int i = 1; i < nums.length; i++) {
            max[i] = min[i] = nums[i];
            if (nums[i] > 0) {
                max[i] = Math.max(max[i], max[i - 1] * nums[i]);
                min[i] = Math.min(min[i], min[i - 1] * nums[i]);
            } else {
                max[i] = Math.max(max[i], min[i - 1] * nums[i]);
                min[i] = Math.min(min[i], max[i - 1] * nums[i]);
            }
            val = Math.max(val, max[i]);
        }
        return val;
    }
}

python

class Solution(object):
    def maxProduct(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if nums is None or len(nums) == 0:
            return 0
        maxArr, minArr = [nums[0]] * len(nums), [nums[0]] * len(nums)
        val = nums[0]
        for i in range(1, len(nums)):
            maxArr[i], minArr[i] = nums[i], nums[i]
            if nums[i] > 0:
                maxArr[i] = max(maxArr[i], maxArr[i - 1] * nums[i])
                minArr[i] = min(minArr[i], minArr[i - 1] * nums[i])
            else:
                maxArr[i] = max(maxArr[i], minArr[i - 1] * nums[i])
                minArr[i] = min(minArr[i], maxArr[i - 1] * nums[i])
            val = max(val, maxArr[i])
        return val