判斷一字符串是否能夠另外一字符串從新排列而成

   1:      class Program

   2:      {

   3:          static void Main(string[] args)

   4:          {

   5:              string s = File.ReadAllText(@"e:\test.txt");

   6:              Program p = new Program();

   7:              bool r=p.IsAnagrams_1(s,s);

   8:              r = p.IsAnagrams_2(s,s);

   9:          }

  10:   

  11:          /// <summary>

  12:          /// Check if the two strings have identical counts for each unique char.

  13:          /// 首先使用框架自帶函數OrderByDescending，稍後會做比較

  14:          /// 由於使用了背後機制爲快速排序，因此複雜度爲O(nLogn)

  15:          /// </summary>

  16:          public bool IsAnagrams_1(string s1, string s2)

  17:          {

  18:              if (string.IsNullOrEmpty(s1) || string.IsNullOrEmpty(s1))

  19:              { 

  20:                  throw new ArgumentException("not accept null or empty string");

  21:              }

  22:              else if (s1.Length != s1.Length)

  23:              {

  24:                  return false;

  25:              }

  26:   

  27:              //OrderByDescending背後用的是快速排序法

  28:              char[] charArray1 = s1.OrderByDescending(a => a).ToArray();

  29:              char[] charArray2 = s2.OrderByDescending(a => a).ToArray();

  30:   

  31:              for (int i = 0; i < charArray1.Length; i++)

  32:              {

  33:                  if (charArray1[i] != charArray2[i])

  34:                  {

  35:                      return false;

  36:                  }

  37:              }

  38:              return true;

  39:          }

  40:   

  41:          /// <summary>

  42:          /// 該方法依舊假設輸入爲ASCII串

  43:          /// 複雜度爲O(n)

  44:          /// </summary>

  45:          public bool IsAnagrams_2(string s1, string s2)

  46:          {

  47:              if (string.IsNullOrEmpty(s1) || string.IsNullOrEmpty(s1))

  48:              {

  49:                  throw new ArgumentException("not accept null or empty string");

  50:              }

  51:              else if (s1.Length != s1.Length)

  52:              {

  53:                  return false;

  54:              }

  55:   

  56:              int[] flags = new int[256];

  57:              for (int i = 0; i < s1.Length; i++)

  58:              {

  59:                  char c = s1[i];

  60:                  if (c > 255)

  61:                  {

  62:                      throw new ArgumentException("not accept non-ASCII string");

  63:                  }

  64:   

  65:                  flags[c]++;

  66:              }

  67:   

  68:              for (int i = 0; i < s2.Length; i++)

  69:              {

  70:                  char c = s2[i];

  71:                  if (c > 255)

  72:                  {

  73:                      throw new ArgumentException("not accept non-ASCII string");

  74:                  }

  75:   

  76:                  if (flags[c] == 0)

  77:                  {

  78:                      return false;

  79:                  }

  80:                  flags[c]--;

  81:              }

  82:   

  83:              return true;

  84:          }

  85:   

  86:          /****

  87:           由於複雜度不在一個等級上，因此在處理超長字符串的時候，2比1要快不少，百萬級別長度字串實驗驗證以下

  88:           可是2由於對輸入要求限制較高，因此影響了其應用，實踐中，若是是超長未知格式字符串，不妨先嚐試用法2試探

  89:           若是ASCII佔大多數，不妨分而治之

  90:           ****/

  91:      }