去除A和B数组中的交集
那位同学碰到新的算法题了。
去除A和B数组中的交集 ,即去除A和B数组中重复次数少的部分如A:111B:11,即去除B中的11
下面是我的解决方案,另希望他最后面试能过:
class Program
{
static void Main(string[] args)
{
int[] ar = { 1, 2, 6, 3, 4, 5 };
int[] ar2 = { 100,100, 3, 2, 6, 100, 9, 3, 4, 7 };
try
{
Console.WriteLine(get2rdMax(ar).ToString());
Console.WriteLine(get2rdMax(ar2).ToString());
}
catch(Exception exc)
{
Console.WriteLine(exc.Message);
}
}
private static int get2rdMax(int[] ar)
{
if (ar.Length < 2)
throw new Exception("no second max!");
int max=ar[0], s_max=ar[0];
for (int i = 1; i < ar.Length; i++)
{
int index = ar[i];
if (index == max)
continue;//重复值问题
if (s_max == ar[0] && max == ar[0])
{
s_max = index;
}
if (index > s_max)
{
s_max = index;
}
if (s_max > max)
{
max += s_max;
s_max = max - s_max;
max -= s_max;
}
}
if (max == s_max)
throw new Exception("no second max!");
else
return s_max;
}
{
static void Main(string[] args)
{
int[] ar = { 1, 2, 6, 3, 4, 5 };
int[] ar2 = { 100,100, 3, 2, 6, 100, 9, 3, 4, 7 };
try
{
Console.WriteLine(get2rdMax(ar).ToString());
Console.WriteLine(get2rdMax(ar2).ToString());
}
catch(Exception exc)
{
Console.WriteLine(exc.Message);
}
}
private static int get2rdMax(int[] ar)
{
if (ar.Length < 2)
throw new Exception("no second max!");
int max=ar[0], s_max=ar[0];
for (int i = 1; i < ar.Length; i++)
{
int index = ar[i];
if (index == max)
continue;//重复值问题
if (s_max == ar[0] && max == ar[0])
{
s_max = index;
}
if (index > s_max)
{
s_max = index;
}
if (s_max > max)
{
max += s_max;
s_max = max - s_max;
max -= s_max;
}
}
if (max == s_max)
throw new Exception("no second max!");
else
return s_max;
}
}