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],

• 常规思路

class Solution {public:    int maxProduct(int A[], int n) {        int product = INT_MIN;        for (int i = 0; i < n; i++)        {            int sub = 1;            for (int j = i; j < n; j++)            {                sub *= A[j];                if (sub > product)                {
                         product = sub;                }            }        }        return product;    }};

上述算法时间复杂度过高，LeetCode返回Time Limit Exceeded错误提示。

• 动态规划思路
  由于数组中有负数存在，一个负数乘以一个负数可能得到一个极大的正数，因此需要维护两个局部变量max_local和min_local。转移方程式如下：
  temp = max_local;
  max_local[i] = max(max(max_local * A[i], min_local * A[i]), A[i]);
  min_local[i] = min(min(temp * A[i], min_local * A[i]), A[i]);
• 实现代码

/*************************************************************    *  @Author   : 楚兴    *  @Date     : 2015/2/8 21:43    *  @Status   : Accepted    *  @Runtime  : 13 ms*************************************************************/class Solution {public:    int maxProduct(int A[], int n) {        if (n == 0)        {            return 0;        }        int max_local = A[0];        int min_local = A[0];        int global = A[0];        for (int i = 1; i < n; i++)        {            int temp = max_local;            max_local = max(max(max_local * A[i], min_local * A[i]), A[i]);            min_local = min(min(temp * A[i], min_local * A[i]), A[i]);            global = max(global, max_local);        }        return global;    }};