【leetcode】 Unique Binary Search Trees (middle)☆

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.

找数字连续最大乘积子序列。

思路：这个麻烦在有负数和0，我的方法，如果有0，一切都设为初始值。

对于两个0之间的数若有奇数个负数，那则有两种情况，第一种是不要第一个负数和之前的值，第二种是不要最后一个负数和之后的值，用negtiveFront和negtiveBack表示。没有负数就是不要第一个负数和之前的值的情况。

int maxProduct(int A[], int n) {
        if(n == )
            return ;

        int MaxAns = A[];
        int negtiveFront = (A[] == ) ?  : A[];
        int negtiveBack = (A[] < ) ?  : ;

        for(int i = ; i < n; i++)
        {
            if(A[i] == )
            {
                MaxAns = (MaxAns > ) ? MaxAns : ;
                negtiveFront = ;
                negtiveBack = ;
            }
            else if(A[i] < )
            {
                negtiveFront *= A[i];
                MaxAns = max(negtiveFront, MaxAns);
                if(negtiveBack == )
                {
                    negtiveBack = ;
                }
                else
                {
                    negtiveBack *= A[i];
                    MaxAns = max(negtiveBack, MaxAns);
                }
            }
            else
            {
                negtiveFront *= A[i];
                negtiveBack *= A[i];
                MaxAns = max(negtiveFront, MaxAns);
                if(negtiveBack > )
                {
                    MaxAns = max(negtiveBack, MaxAns);
                }

            }
        }

        return MaxAns;
    }

答案的思路：同时维护包括当前数字A[k]的最大值f(k)和最小值g(k)

f(k) = max( f(k-1) * A[k], A[k], g(k-1) * A[k] )
g(k) = min( g(k-1) * A[k], A[k], f(k-1) * A[k] )

再用一个变量Ans存储所有f(k)中最大的数字就可以了

int maxProduct2(int A[], int n) {
        if(n == )
            return ;

        int MaxAns = A[]; //包括当前A【i】的连续最大乘积
        int MinAns = A[]; //包括当前A【i】的连续最小乘积
        int MaxSoFar = A[]; //整个数组的最大乘积

        for(int i = ; i < n; i++)
        {
            int MaxAnsTmp = MaxAns;
            int MinAnsTmp = MinAns;
            MaxAns = max(MaxAnsTmp * A[i], max(MinAnsTmp * A[i], A[i]));
            MinAns = min(MinAnsTmp * A[i], min(MaxAnsTmp * A[i], A[i]));
            MaxSoFar = max(MaxSoFar, MaxAns);

        }

        return MaxSoFar;
    }