LeetCode Maximum Product Subarray

class Solution {

public:

    int maxProduct(int A[], int n) {

        if (A == NULL || n < 1) {

            return INT_MIN;

        }

        int res = INT_MIN;

        int tr = 1;

        for (int i=0; i<n; i++) {

            tr = tr * A[i];

            if (tr > res) {

                res = tr;

            }

            if (tr == 0) {

                tr = 1;

            }

        }

        tr = 1;

        for (int i=n-1; i>=0; i--) {

            tr = tr * A[i];

            if (tr > res) {

                res = tr;

            }

            if (tr == 0) {

                tr = 1;

            }

        }

        return res;

    }

};

 更叼的方法，不过速度上没有明显提升

class Solution {

public:

    int maxProduct(int A[], int n) {

        // store the result that is the max we have found so far

        int r = A[0];

    

        // imax/imin stores the max/min product of

        // subarray that ends with the current number A[i]

        for (int i = 1, imax = r, imin = r; i < n; i++) {

            // multiplied by a negative makes big number smaller, small number bigger

            // so we redefine the extremums by swapping them

            if (A[i] < 0)

                swap(imax, imin);

    

            // max/min product for the current number is either the current number itself

            // or the max/min by the previous number times the current one

            imax = max(A[i], imax * A[i]);

            imin = min(A[i], imin * A[i]);

    

            // the newly computed max value is a candidate for our global result

            r = max(r, imax);

        }

        return r;

    }

};

 第二轮：

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.

还是中间的方法更好一些

class Solution {

public:

    int maxProduct(int A[], int n) {

        int maxv = A[0];

        int minv = A[0];

        int maxt = A[0];

        for (int i=1; i<n; i++) {

            int mxv = maxv;

            int miv = minv;

            maxv = max(max(mxv * A[i], miv * A[i]), A[i]);

            minv = min(min(miv * A[i], mxv * A[i]), A[i]);

            maxt = max(maxv, maxt);

        }

        return maxt;

    }

};