[英]create asterisk tree with C#
大家好我試着解決星號樹問題,發現我的代碼工作不正常,可以改進。
這是預期的輸出
輸入:5
*
* * *
* * * * *
* * *
*
輸入:4
* * * *
* *
* * * *
這是我的代碼
static void Main(string[] args)
{
Console.Write("input:");
char input = Console.ReadKey().KeyChar;
if (char.IsDigit(input))
{
int couter = (int)char.GetNumericValue(input);
Console.WriteLine();
if (couter % 2 != 0)
{
for (int i = 1; i <= couter; i++)
{
for (int j = 3; j > i; j--)
{
Console.Write(" ");
}
for (int k = 1; k <= i; k++)
{
Console.Write(" *");
}
Console.WriteLine();
}
for (int i = couter - 1; i >= 3; i--)
{
for (int j = 1; j <= i; j++)
{
if (j <= couter - i)
{
Console.Write(" ");
}
else
{
Console.Write("* ");
}
}
Console.WriteLine();
}
}
else
{
for (int i = couter; i > 3; i--)
{
for (int j = 1; j <= i; j++)
{
if (couter - i >= j)
{
Console.Write(" ");
}
else
{
Console.Write("* ");
}
}
Console.WriteLine();
}
for (int i = couter - 1; i <= couter; i++)
{
for (int j = 0; j < i; j++)
{
Console.Write("* ");
}
Console.WriteLine();
}
}
}
}
請你幫我解決這個問題。
最近,我認為我在算法方面很差,而且有點復雜的問題。 有人知道有用的鏈接或我如何提高這項技能,請告訴我。
謝謝,
檢查此頁面輸入5(菱形): http : //www.dreamincode.net/forums/topic/126715-diamond-asterisk/
我已將其翻譯為C# - 現在它顯示的鑽石大小可以在變量'rows'中設置:
int rows = 5;
StringBuilder sb = new StringBuilder();
// top part
for (int i = 1; i <= rows; i++)
{
for (int j = 1; j <= rows - i; j++)
sb.Append(' ');
for (int k = 1; k <= 2 * i - 1; k++)
sb.Append('*');
sb.AppendLine();
}
//bottom part
for (int n = rows - 1; n > 0; n--)
{
for (int l = 1; l <= rows - n; l++)
sb.Append(' ');
for (int m = 1; m <= 2 * n - 1; m++)
sb.Append('*');
sb.AppendLine();
}
Console.Write(sb.ToString());
我最初不願發布它,因為它絕對聞起來像家庭作業......
無論如何,這是一段工作代碼:
static void Main(string[] args)
{
Console.Write("input:");
char input = Console.ReadKey().KeyChar;
if (char.IsDigit(input))
{
int couter = (int)char.GetNumericValue(input);
Console.WriteLine();
if (couter % 2 != 0)
PrintDiamond(couter);
else
PrintHourGlass(couter);
}
Console.ReadLine();
}
private static void PrintDiamond(int couter)
{
bool moreAsterisks = true;
for (int row = 0; row < couter; row++)
{
int nAsterisks = moreAsterisks ? (2 * row) + 1 : 2 * (couter - row - 1) + 1;
int nSpaces = (couter - nAsterisks) / 2;
if (row == (couter - 1) / 2)
moreAsterisks = false;
for (int i = 0; i < nSpaces; i++)
Console.Write(" ");
for (int i = 0; i < nAsterisks; i++)
Console.Write("*");
for (int i = 0; i < nSpaces; i++)
Console.Write(" ");
Console.WriteLine();
}
}
private static void PrintHourGlass(int couter)
{
bool moreAsterisks = false;
for (int row = 0; row < couter - 1; row++)
{
int nAsterisks = moreAsterisks ? couter - 2 * (couter - row - 2) : couter - (2 * row);
int nSpaces = (couter - nAsterisks) / 2;
if (row == (couter - 2) / 2)
moreAsterisks = true;
for (int i = 0; i < nSpaces; i++)
Console.Write(" ");
for (int i = 0; i < nAsterisks; i++)
Console.Write("*");
for (int i = 0; i < nSpaces; i++)
Console.Write(" ");
Console.WriteLine();
}
}
PS
:它適用於任何數字,而不僅僅是4-5 ......
這是針對您的問題的縮小LINQ解決方案:
class Program
{
static void Main(string[] args)
{
Console.Write("input: ");
string line = Console.ReadLine();
int n;
if (!int.TryParse(line, out n))
{
Console.WriteLine("Enter a valid integer number.");
return;
}
for (int i = 0; i < n; i++)
{
int l = Math.Abs(n - i * 2 - 1) + 1;
if (n % 2 != 0) l = n - l + 1;
Console.Write(Enumerable.Repeat(" ", n - l).DefaultIfEmpty("").Aggregate((a, b) => a + b));
Console.WriteLine(Enumerable.Repeat("* ", l).DefaultIfEmpty("").Aggregate((a, b) => a + b));
}
}
}
這很簡單; 在循環內部,如果輸入是偶數,則第一行計算第i個菱形行的長度,第二行校正奇數輸入的計算。 其余兩行使用一些LINQ技巧打印第i行。 如果您不喜歡使用LINQ進行炫耀,可以使用常規for循環替換thoose行(這可能會更快)。 然后代碼看起來像:
class Program
{
static void Main(string[] args)
{
Console.Write("input: ");
string line = Console.ReadLine();
int n;
if (!int.TryParse(line, out n))
{
Console.WriteLine("Enter a valid integer number.");
return;
}
for (int i = 0; i < n; i++)
{
int l = Math.Abs(n - i * 2 - 1) + 1;
if (n % 2 != 0) l = n - l + 1;
//Console.Write(Enumerable.Repeat(" ", n - l).DefaultIfEmpty("").Aggregate((a, b) => a + b));
//Console.WriteLine(Enumerable.Repeat("* ", l).DefaultIfEmpty("").Aggregate((a, b) => a + b));
for (int c = 0; c < n - l; c++) Console.Write(" ");
for (int c = 0; c < l; c++) Console.Write("* ");
Console.WriteLine();
}
}
}
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