LeetCode--Maximum Product Subarray

标签：leetcode   动态规划   

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.

class Solution 
{
public:
    int maxProduct(int A[], int n) 
    {
        if(n==0)
            return -1;
        if(n==1)
            return A[0];
        int max = 0;
        int min = 0;
        int res = 0;
        if(A[0]>0)
            max = A[0];
        else 
            min = A[0];
        res = max;
        int temp_max;
        int temp_min;
        for(int i=1; i<n; i++)
        {
            temp_max = max;
            temp_min = min;
            if(A[i] > 0)
            {
                max = (temp_max!=0)? (temp_max*A[i]):A[i];
                min = temp_min*A[i];
            }
            else if(A[i] < 0)
            {
                min = (temp_max!=0) ? (temp_max*A[i]):A[i];
                max = temp_min*A[i];
            }
            else
            {
                max = 0;
                min = 0;
            }
            res = (max>res)? max:res;
        }
        return res;
    }
};

另一种方法：

class Solution 
{
public:
    int maxProduct(int A[], int n) 
    {
        if(n==0)
            return -1;
        if(n==1)
            return A[0];
        int* max = new int[n];
        int* min = new int[n];
        for(int i=0; i<n; i++)
        {
            max[i] = 0;
            min[i] = 0;
        }
        if(A[0]>0)
            max[0] = A[0];
        else 
            min[0] = A[0];
        for(int i=1; i<n; i++)
        {
            if(A[i]>0)
            {
                if(max[i-1]!=0)
                    max[i] = max[i-1]*A[i];
                else
                    max[i] = A[i];
                min[i] = min[i-1]*A[i];
               
            }
            else if(A[i]<0)
            {
                max[i] = min[i-1]*A[i];
                if(max[i-1]!=0)
                    min[i] = max[i-1]*A[i];
                else
                    min[i] = A[i];
            }
        }
        int res = max[0];
        for(int i=0; i<n; i++)
        {
            if(max[i]>res)
                res = max[i];
        }
        return res;
    }
};

LeetCode--Maximum Product Subarray

标签：leetcode   动态规划   

原文地址：http://blog.csdn.net/shaya118/article/details/42746151