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.

第一印象是DP，不过用了个比较啰嗦的解法

题目中是整数类型，就比较简单了，把数组按0分开，子数组的最大乘积肯定就是最大乘积了

class Solution {
public:
    int product(vector<int> & arr,int s,int e){
        int sum = INT_MIN;
        if(s<e) {
            sum = 1;
            for(int i = s;i<e;++i){
                sum *= arr[i];
            }
        }
        return sum;
    }

    int calcMax( vector<int> & arr,int s,int e) {
        vector<int> v(e-s);
        int sum = INT_MIN;
        copy(arr.begin()+s,arr.begin()+e,v.begin());
        int negativCount = count_if(v.begin(),v.end(),[](int x){return x<0;});
        if(negativCount%2==0){
            //有偶数个负数直接得乘积
            sum = product(v,0,v.size());
        }else{
            //奇数个负数分别算下两边的最大值
            int l =  find_if(v.begin(),v.end(),[](int x){return x<0;}) - v.begin();
            int lmax = max(product(v,0,l),product(v,l+1,v.size()));
            int r = find_if(v.rbegin(),v.rend(),[](int x){return x<0;}).base() - v.begin()-1;
            int rmax = max(product(v,0,r),product(v,r+1,v.size()));
            sum = max(lmax,rmax);
        }
        return sum;
    }

    int maxProduct(int A[], int n) {
        vector<int> arr(A,A+n);
        vector<int> val;
        //将数组按0分割成子数组，分别计算每个子数组的最大乘积
        int s = -1 , e = 0;
        while(true) {
            if(arr[e]==0 || e==arr.size()) {
                if(s+1 != e)
                    val.push_back(calcMax(arr,s+1,e));
                s = e;
            }
            ++e;
            if(e>arr.size())break;
        }
        //防止数组只有一个元素的情况
        val.push_back(*max_element(arr.begin(),arr.end()));
        return *max_element(val.begin(),val.end());
    }
};