算法大比拼(看看谁是真真的高手)?(java C# C++ C的高手请进)
求算一任意长度字符串中不同的字符以及它的个数?(设计一个漂亮的算法,快而优美)
如在 字符串"abcdefgabc"中求算出的结果是 
a,2
b,2
c,2
d,1
e,1
f,1
g,1

以一个普通的算法开个头吧.
 public static class MyMethod
  {
  public static Queue<char> GetOnlyCharFromString(string str)
  {
  Queue<char> myQueue = new Queue<char>();
  for (int i = 0; i < str.Length; i++)
  {
  if (myQueue.Contains(str[i]) == false)
  myQueue.Enqueue(str[i]);
  }
  return myQueue;
  }
  public static Dictionary<char,int> GetInfoFormString(string str)
  {
  Queue<char> myQueue = GetOnlyCharFromString(str );
  Dictionary<char,int> myDictionary=new Dictionary<char,int>();
  int oldQueueCount = myQueue.Count;//记录下它原有的量;
  for(int i=0;i<oldQueueCount ;i++)
  {
  char myChar=myQueue.Dequeue();
  int num=0;
  for(int j=0;j<str.Length;j++)
  {
  if(str[j]==myChar)
  num++;
  }
  myDictionary.Add(myChar,num);
  }
  return myDictionary;
  }

  }

------解决方案--------------------
不知道效率怎么样
但是这是一个思路
你可以用正则表达式去匹配
从ASCII为0的到128
以下是.NET下正则表达式的使用方法
http://msdn.microsoft.com/zh-cn/library/system.text.regularexpressions.regex(VS.80).aspx
// Define a regular expression for repeated words.
Regex rx = new Regex(@"\b(?<word>\w+)\s+(\k<word>)\b",
RegexOptions.Compiled | RegexOptions.IgnoreCase);

// Define a test string.
string text = "The the quick brown fox fox jumped over the lazy dog dog.";

// Find matches.
MatchCollection matches = rx.Matches(text);

// Report the number of matches found.
Console.WriteLine("{0} matches found.", matches.Count);

// Report on each match.
foreach (Match match in matches)
{
string word = match.Groups["word"].Value;
int index = match.Index;
Console.WriteLine("{0} repeated at position {1}", word, index);
}

------解决方案--------------------

C# code

            string str = "abcdefgabc";
            var result = from c in str
                         group c by c into g
                         select g;
            foreach (var group in result)
                Console.WriteLine(group.Key + "," + group.Count());

------解决方案--------------------
C++ Code
int main(void)
{	
	string str = "abcdeffasljdlfjaslflsadjflkasdfjlsadlfgabc";
	sort(str.begin(), str.end());	//先排序
	
	cout<<str<<endl;
	char c;
	int count;

	int i = 0;
	while(i < str.length()){	
		c = str[i];
		count = 1;
		while(c == str[++i]){
			count++;
		}
		cout<<c<<","<<count<<endl;
	}
	return 0;
}

------解决方案--------------------
SQL code

declare @a varchar(max) 
set @a='abcdefgabc'
declare @b table(s varchar(1))
declare @n int
declare @i int
set @i=1
set @n=len(@a)
set rowcount @n
while @i<=@n
begin
    insert into @b select substring(@a,@i,1)
    set @i=@i+1
end
set rowcount 0
select count(s),s from @b group by s