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;
    }
};