LeetCode 152. 乘积最大子数组

动态规划

1. dp_max[i]表示以第i个元素结尾的最大乘积
2. dp_min[i]表示以第j个元素结尾的最小乘积（考虑负数情况）
3. dp_max[i] = max{dp_max[i-1] * nums[i], dp_min[i-1] * nums[i], nums[i]};
4. dp_min[i] = min{dp_min[i-1] * nums[i], dp_max[i-1] * nums[i], nums[i]};

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        
        int n = nums.size();
        long long dp_max = nums[0];
        long long dp_min = nums[0];

        long long res = nums[0];
        for(int i=1; i<n; i++)
        {
            int tmp_max = max(dp_max * nums[i] , max((long long)nums[i], dp_min * nums[i]));
            int tmp_min = min(dp_min * nums[i] , min((long long)nums[i], dp_max * nums[i]));
            dp_max = tmp_max;
            dp_min = tmp_min;
            
            res = max(res,dp_max);
        }
        return res;
    }
};