这个方法相对比较冗长,因为它需要检查每一行两次...
public static string[,] RemoveEmptyRows(string[,] strs)
{
int length1 = strs.GetLength(0);
int length2 = strs.GetLength(1);
int nonEmpty = 0;
for (int i = 0; i < length1; i++)
{
for (int j = 0; j < length2; j++)
{
if (strs[i, j] != null)
{
nonEmpty++;
break;
}
}
}
string[,] strs2 = new string[nonEmpty, length2];
for (int i1 = 0, i2 = 0; i2 < nonEmpty; i1++)
{
for (int j = 0; j < length2; j++)
{
if (strs[i1, j] != null)
{
for (int k = 0; k < length2; k++)
{
strs2[i2, k] = strs[i1, k];
}
i2++;
break;
}
}
}
return strs2;
}
使用方法如下:
string[,] options = new string[100, 3];
options[1, 0] = "Foo";
options[3, 1] = "Bar";
options[90, 2] = "fiz";
options = RemoveEmptyRows(options);
正如Alexei所建议的,还有另一种方法可以做到这一点:
public static string[,] RemoveEmptyRows2(string[,] strs)
{
int length1 = strs.GetLength(0);
int length2 = strs.GetLength(1);
// First we put somewhere a list of the indexes of the non-emtpy rows
var nonEmpty = new List<int>();
for (int i = 0; i < length1; i++)
{
for (int j = 0; j < length2; j++)
{
if (strs[i, j] != null)
{
nonEmpty.Add(i);
break;
}
}
}
// Then we create an array of the right size
string[,] strs2 = new string[nonEmpty.Count, length2];
// And we copy the rows from strs to strs2, using the nonEmpty
// list of indexes
for (int i1 = 0; i1 < nonEmpty.Count; i1++)
{
int i2 = nonEmpty[i1];
for (int j = 0; j < length2; j++)
{
strs2[i1, j] = strs[i2, j];
}
}
return strs2;
}
在时间和内存的权衡中,这个选择了时间。它可能更快,因为它不必两次检查每一行,但它使用了更多的内存,因为它在某个地方放置了一个非空索引列表。
List<Tuple<string,string,string>>
,或者是一个包含三个字符串属性的自定义类实例的列表。 - juharr