求子数组的最大和

方案一：遍历所有子数组，复杂度o(n^3)

    public static int Max1()
        {
            int[] array = { 1, -2, 3, 10, -4, 7, 2, -5 };
            int length = array.Length;
            int maximum = array[0];
            int sum = 0;
            for (int i = 0; i < length; i++) //一二两层循环用来遍历子数组
            {
                for (int j = i; j < length; j++)  
                {
                    for (int k = i; k <= j; k++) //第三层循环用来计算子数组的和
                    {
                        sum += array[k];
                    }
                    if (sum > maximum)
                        maximum = sum;
                    sum = 0;  
                }
            }
            return maximum;
        }

方案二：抛弃掉临时累加和为负数的子序列，复杂度o(n)

 public static int Max2() 
        {
            int[] array = { 1, -2, 3, 10, -4, 7, 2, -5 };
            int length = array.Length;
            int maximum =array[0];
            int sum = 0;

            for (int i = 0; i < length; i++)
            {
                if (sum < 0)
                    sum = array[i]; //和为负数，后面无论是正数还是负数都需要被覆盖掉
                else
                    sum += array[i];
                
                if (maximum < sum)
                    maximum = sum;
            }

            return maximum;
        }