学步园

数组为：   int[] args = { -2, 11, -4, 13, -5, 2, -5, -3, 12, -9 };

最大和为：21

分治(devide-and-conquer):

      其想法是把问题分成两个大致相等的子问题,然后递归地对他们求解,这是“分”的部分。“治”阶段将两个子问题的解修补到一起并可能再做些少量的附加工作,最后得到整个问题的解。

     对于上面的测试数据被分成两组 {-2, 11, -4, 13，-5} ;{ 2, -5, -3, 12, -9}。最终结果存在种可能:在left数组中;在right数组中;找到包含

left最后一个元素（在例子中是
-5
）在


left
的最大子序列（例子中是


11, -4, 13，-5）和包含

right起始元素（2
）在
Second
的最大子序列（

2, -5, -3, 12）然后相加的结果。

  下面我把我简要写的C#测试代码粘贴如下：

  
        /// <summary>
        /// 分治 算法
        /// </summary>
        /// <param name="a">A.</param>
        /// <param name="left">The left.</param>
        /// <param name="right">The right.</param>
        /// <returns></returns>
        int getMaxSumUsingDivideConquer(int[] a, int left, int right)
        {
            if (left == right)
            {
                if (args[left] > 0)
                {
                    return args[left];
                }
                else
                {
                    return 0;
                }
            }

            int center = (left + right) / 2;

            int maxLeftSum = getMaxSumUsingDivideConquer(a, left, center);

            int maxRightSum = getMaxSumUsingDivideConquer(a, center + 1, right);

            int maxLeftBorderSum = 0, leftBorderSum = 0;
            for (int i = center; i >= left; i--)
            {
                leftBorderSum += a[i];

                if (leftBorderSum > maxLeftBorderSum)
                {
                    maxLeftBorderSum = leftBorderSum;
                }
            }

            int maxRightBorderSum = 0, RightBorderSum = 0;
            for (int i = center + 1; i <= right; i++)
            {
                RightBorderSum += a[i];

                if (RightBorderSum > maxRightBorderSum)
                {
                    maxRightBorderSum = RightBorderSum;
                }
            }

            return GetMaxValueFromTreeNumber(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);       //   判断三个数最大值的函数

        }

        /// <summary>
        /// Gets the max value from tree number.
        /// </summary>
        /// <param name="maxLeftSum">The max left sum.</param>
        /// <param name="maxRightSum">The max right sum.</param>
        /// <param name="third">The third.</param>
        /// <returns></returns>
        private int GetMaxValueFromTreeNumber(int maxLeftSum, int maxRightSum, int third)
        {
            int max = 0;
            if (max < maxLeftSum)
            {
                max = maxLeftSum;
            }
            if (max < maxRightSum)
            {
                max = maxRightSum;
            }
            if (max < third)
            {
                max = third;
            }

            return max;
        }

具体调用方法如下：

int total = getMaxSumUsingDivideConquer(args, 0, args.Length - 1);

MessageBox.Show("the value of maxSum is:" + total.ToString());