两道MS的面试题 及答案

1.一个整型数组,知道长度,里面存放有:负数/正数/零,请求"子串之和最大"的子串,函数返回该子串之和.函数形式:   int   f(int*   array,   int   length);  
          举例:   array   =   {7,6,3,-10,2,-8,6}  
                      和最大的子串为:   7   +   6   +   3   =   16  
                      函数返回:   16  
          建议测试用例:   {7,6,3,-10,200,-8,6}   ->   return   7   +   6   +   3   +   (-10)   +   200   =   206,  
                                      {7,6,3,-100,2,-8,6}   ->   return   2,  
                                      {100,-200,100,-10,100}   ->   return   100   +   (-10)   +   100   =   190.  
          其他情况可另行测试.  

答案如下:

        static int GetMaxString(int[] arr)
        {
            int curMax = 0;

            int i = 0;
            int j = 0;

            int length = arr.Length;

            int[] arrMax = new int[length];

            for (; i < arr.Length; i++)
            {
                if (arr[i] < 0)
                    continue;
                else
                    break;
            }

            if (i == arr.Length && arr[i - 1] < 0)
            {
                int tmpMax2 = arr[0];

                for (int m = 0; m < arr.Length; m++)
                {
                    if (tmpMax2 < arr[m])
                    {
                        tmpMax2 = arr[m];
                    }
                }

                return tmpMax2;
            }

            for (; i < arr.Length; i++)
            {
                if (arr[i] >= 0)
                {
                    curMax += arr[i];

                    if (i == arr.Length - 1)
                    {
                        arrMax[j++] = curMax;
                    }
                }
                else
                {
                    arrMax[j++] = curMax;

                    if (curMax + arr[i] >= 0)
                    {
                        curMax += arr[i];
                    }
                    else
                    {
                        curMax = 0;
                    }
                }
            }

            int tmpMax = arrMax[0];

            for (int k = 1; k < j; k++)
            {
                if (tmpMax < arrMax[k])
                {
                    tmpMax = arrMax[k];
                }
            }

            return tmpMax;
        }

2.一个没有排序的链表,比如list={a,l,x,b,e,f,f,e,a,g,h,b,m},请去掉"重复项,并保留原顺序",以上链表去掉重复项后为newlist={a,l,x,b,e,f,g,h,m},请写出一个高效算法(时间比空间更重要)  
          提示:如果采用先排序,在去掉重复项的办法,复杂度为n+nlogn(采用快排),显然不是最优解,而且破还了"原顺序".还有其他解法.  

答案如下:

        static void Main(string[] args)
        {
            Link link1 = new Link('a');

            Link link2 = new Link('l');
            Link link3 = new Link('x');
            Link link4 = new Link('b');
            Link link5 = new Link('e');

            Link link6 = new Link('f');
            Link link7 = new Link('f');
            Link link8 = new Link('e');
            Link link9 = new Link('a');

            Link link10 = new Link('g');
            Link link11 = new Link('h');
            Link link12 = new Link('b');
            Link link13 = new Link('m');

            link1.Next = link2;
            link2.Next = link3;
            link3.Next = link4;
            link4.Next = link5;
            link5.Next = link6;
            link6.Next = link7;
            link7.Next = link8;
            link8.Next = link9;
            link9.Next = link10;
            link10.Next = link11;
            link11.Next = link12;
            link12.Next = link13;
            link13.Next = null;

            link1.Prev = null;
            link2.Prev = link1;
            link3.Prev = link2;
            link4.Prev = link3;
            link5.Prev = link4;
            link6.Prev = link5;
            link7.Prev = link6;
            link8.Prev = link7;
            link9.Prev = link8;
            link10.Prev = link9;
            link11.Prev = link10;
            link12.Prev = link11;
            link13.Prev = link12;

            Link link = link1;


            int[] tmp = new int[256];

            Link tmpLink = link;

            while (tmpLink != null)
            {

                tmp[Convert.ToInt32(tmpLink.Chr)]++;

                tmpLink = tmpLink.Next;
            }

            Link tmpLink2 = link;
            int count = 0;

            while (tmpLink2 != null)
            {
                count++;

                if (tmp[Convert.ToInt32(tmpLink2.Chr)] == 2)
                {
                    tmp[Convert.ToInt32(tmpLink2.Chr)]++;
                }
                else if (tmp[Convert.ToInt32(tmpLink2.Chr)] == 3)
                {
                    tmpLink2.Prev.Next = tmpLink2.Next;
                }

                tmpLink2 = tmpLink2.Next;
            }


            Link tmpLink3 = link;
            while (tmpLink3 != null)
            {
                Console.WriteLine(tmpLink3.Chr.ToString());

                tmpLink3 = tmpLink3.Next;
            }

            Console.Read();
        }