子数组最大累加和

---

/** @brief
 * 给定一个数组arr，返回子数组的最大累加和
 * arr=[1,-2,3,5,-2,6,-1]，所有的子数组中，[3,5,-2,6]可以累加出最大的和12
 *
 * @param arr[] int
 * @param length int
 * @return int
 *
 */
int MaxSubSum(int arr[], int length)
{
    int ret = INT_MIN;
    int sum = 0;
    for(int i = 0; i < length; ++i)
    {
        sum += arr[i];
        ret = max(sum, ret);
        if(sum < 0) sum = 0;
    }

    return ret;
}

---

/** @brief
 * 找到两个不相容子数组的最大和(子数组不能为空)
 * 不相容，即下标区间不重叠
 * 数组[1,-1,0,-2,3,5,-2,8,7,-4]，两个不相容子数组分别为[3,5]和[8,7]时累加和最大
 * 数组[3,-1,0,-2,3,5,-2,8,7,-4]，两个不相容子数组分别为[3]和[3,5,-2,8,7]时累加和最大
 *
 * @param arr[] int
 * @param length int
 * @return int
 *
 */
int TwoSubarrayMaxSum(int arr[], int length)
{
    int *pSums = new int[length];

    int maxSum = INT_MIN;
    int sum = 0;
    for(int i = 0; i < length; ++i)
    {
        sum += arr[i];
        maxSum = max(maxSum, sum);
        pSums[i] = maxSum;
        if(sum < 0) sum = 0;
    }

    int ret = INT_MIN;
    sum = 0;
    maxSum = INT_MIN;
    for(int i = length - 1; i > 0; --i)
    {
        sum += arr[i];
        maxSum = max(maxSum, sum);
        pSums[i] = maxSum;
        if(sum < 0) sum = 0;
        ret = max(ret, pSums[i] + pSums[i - 1]);

    }

    delete[] pSums;

    return ret;
}

---

/** @brief
 * 子矩阵的最大累加和问题
 * 给定一个矩阵matrix，其中的值有正、有负、有0，返回子矩阵的最大累加和。
 *
 * @param matrix vector<vector<int> > -矩阵
 * @param row int -矩阵行数
 * @param column int -矩阵列数
 * @return int
 *
 */
int MaxSubMatrixSum(vector<vector<int> > matrix, int row, int column)
{
    int *pNums = new int[column];
    int ret = INT_MIN;

    for(int i = 0; i < row; ++i)
    {
        memset(pNums, 0, sizeof(int) * column);
        for(int j = i; j < row; ++j)
        {
            for(int col = 0; col < column; ++col)
            {
                pNums[col] += matrix[j][col];
            }

            ret = max(ret, MaxSubSum(pNums, column));
        }
    }

    delete[] pNums;
    return ret;
}