Given an array of integers nums, find the maximum length of a subarray where the product of all its elements is positive.
A subarray of an array is a consecutive sequence of zero or more values taken out of that array.
Return the maximum length of a subarray with positive product.

状态转移方程
1.nums[i] > 0
pos[i] = pos[i-1] + 1;
neg[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1
2.nums[i] < 0
neg[i] = pos[i-1] + 1;
pos[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1;
3.nums[i] == 0
neg[i] = 0;
pos[i] = 0;

边界条件
if(nums[0] > 0) pos[0] = 1
else neg[0] = 1

代码

class Solution {
public:
    int getMaxLen(vector<int>& nums) {
        int len = nums.size();
        int pos[100005] = {0}, neg[100005] = {0};
        if(nums[0] > 0)
        pos[0] = 1;
        else if(nums[0] < 0)
        neg[0] = 1;

        int maxl = pos[0];
        for(int i = 1; i < len; i++)
        {
            if(nums[i] < 0)
            {
                neg[i] = pos[i-1] + 1;
                pos[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1;
            }
            else if(nums[i] == 0)
            {
                neg[i] = 0;
                pos[i] = 0;
            }
            else if(nums[i] > 0)
            {
                pos[i] = pos[i-1] + 1; 
                neg[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1;
            }
            if(pos[i] > maxl)
            maxl = pos[i];
        }
        return maxl;
    }
};