刷题152. Maximum Product Subarray

一、题目说明

题目152. Maximum Product Subarray，给一列整数，求最大连续子序列，其乘积最大。难度是Medium！

二、我的解答

这个题目，用双重循环就可以了。

class Solution{
	public:
		int maxProduct(vector<int>& nums){
			if(nums.size()<=1) return nums[0];
			product = INT_MIN;
			int len = nums.size();
			
			for(int i=0;i<len;i++){
				int p = nums[i];
				if(p>product) product = p;
				for(int t=i+1;t<len;t++){
					p *= nums[t];
					if(p>product) product = p;
				}
			}
			return product;
		}
	private:
		int product;
};

性能如下：

Runtime: 200 ms, faster than 6.16% of C++ online submissions for Maximum Product Subarray.
Memory Usage: 9.1 MB, less than 82.50% of C++ online submissions for Maximum Product Subarray.

三、优化措施

仔细再读读题目，一列整数，上述方法太“通用”，一次循环就可以了。

class Solution{
	public:
		//dp，其中dp[i]表示以第i个元素结尾的最大乘积 
		int maxProduct(vector<int>& nums){
			if(nums.size()<=1) return nums[0];
			dpMax = nums[0];
			dpMin = nums[0];
			maxProd = nums[0];
			int len = nums.size();
			
			for(int i=1;i<len;i++){
				int preMax = dpMax;
				dpMax = max(dpMin*nums[i],max(nums[i],dpMax*nums[i]));
				dpMin = min(dpMin*nums[i],min(preMax*nums[i],nums[i]));
				cout<<"i="<<i<<",dpMax="<<dpMax<<",dpMin="<<dpMin<<"
"; 
				maxProd = max(dpMax,maxProd);
			}
			return maxProd;
		}
	private:
		int dpMax;
		int dpMin;
		int maxProd;
};

Runtime: 16 ms, faster than 9.45% of C++ online submissions for Maximum Product Subarray.
Memory Usage: 9.1 MB, less than 75.00% of C++ online submissions for Maximum Product Subarray.