C#，回文分割问题（Palindrome Partitioning Problem）算法与源代码

1 回文串

“回文串”是一个正读和反读都一样的字符串，初始化标志flag=true，比如“level”或者“noon”等等就是回文串。

2 回文分割问题

给定一个字符串，如果该字符串的每个子字符串都是回文的，那么该字符串的分区就是回文分区。
例如，“aba | b | bbabb | a | b | aba”是“abababababa”的回文分区。
确定给定字符串的回文分区所需的最少切割。
例如，“ababababababa”至少需要3次切割。
这三个分段是“a | babbab | b | ababa”。
如果字符串是回文，则至少需要0个分段。
如果一个长度为n的字符串包含所有不同的字符，则至少需要n-1个分段。

3 源程序

using System;
using System.Collections;
using System.Collections.Generic;

namespace Legalsoft.Truffer.Algorithm
{
   public static partial class Algorithm_Gallery
   {
#region 算法1
private static bool PPP_IsPalindrome(string String, int i, int j)
       {
           while (i < j)
           {
               if (String[i] != String[j])
               {
                   return false;
               }
               i++;
               j--;
           }
           return true;
       }

       public static int Min_Palindrome_Partion(string str, int i, int j)
       {
           if (i >= j || PPP_IsPalindrome(str, i, j))
           {
               return 0;
           }
           int ans = Int32.MaxValue, count;
           for (int k = i; k < j; k++)
           {
               count = Min_Palindrome_Partion(str, i, k) + Min_Palindrome_Partion(str, k + 1, j) + 1;

               ans = Math.Min(ans, count);
           }
           return ans;
       }
#endregion

#region 算法2
public static int Min_Palindrome_Partion(string str)
       {
           int n = str.Length;
           int[,] C = new int[n, n];
           bool[,] P = new bool[n, n];

           int i, j, k, L;
           for (i = 0; i < n; i++)
           {
               P[i, i] = true;
               C[i, i] = 0;
           }

           for (L = 2; L <= n; L++)
           {
               for (i = 0; i < n - L + 1; i++)
               {
                   j = i + L - 1;
                   if (L == 2)
                   {
                       P[i, j] = (str[i] == str[j]);
                   }
                   else
                   {
                       P[i, j] = (str[i] == str[j]) && P[i + 1, j - 1];
                   }
                   if (P[i, j] == true)
                   {
                       C[i, j] = 0;
                   }
                   else
                   {
                       C[i, j] = int.MaxValue;
                       for (k = i; k <= j - 1; k++)
                       {
                           C[i, j] = Math.Min(C[i, j], C[i, k] + C[k + 1, j] + 1);
                       }
                   }
               }
           }
           return C[0, n - 1];
       }
#endregion

#region 算法3
public static int Min_Cutting(string a)
       {
           int[] cut = new int[a.Length];
           bool[,] palindrome = new bool[a.Length, a.Length];
           for (int i = 0; i < a.Length; i++)
           {
               int minCut = i;
               for (int j = 0; j <= i; j++)
               {
                   if (a[i] == a[j] && (i - j < 2 || palindrome[j + 1, i - 1]))
                   {
                       palindrome[j, i] = true;
                       minCut = Math.Min(minCut, j == 0 ? 0 : (cut[j - 1] + 1));
                   }
               }
               cut[i] = minCut;
           }
           return cut[a.Length - 1];
       }
#endregion

#region 算法4
public static int Min_Palindrome_Partion_Second(string str)
       {
           int n = str.Length;
           int[] C = new int[n];
           bool[,] P = new bool[n, n];

           int i, j, L;
           for (i = 0; i < n; i++)
           {
               P[i, i] = true;
           }

           for (L = 2; L <= n; L++)
           {
               for (i = 0; i < n - L + 1; i++)
               {
                   j = i + L - 1;
                   if (L == 2)
                   {
                       P[i, j] = (str[i] == str[j]);
                   }
                   else
                   {
                       P[i, j] = (str[i] == str[j]) && P[i + 1, j - 1];
                   }
               }
           }

           for (i = 0; i < n; i++)
           {
               if (P[0, i] == true)
               {
                   C[i] = 0;
               }
               else
               {
                   C[i] = int.MaxValue;
                   for (j = 0; j < i; j++)
                   {
                       if (P[j + 1, i] == true && 1 + C[j] < C[i])
                       {
                           C[i] = 1 + C[j];
                       }
                   }
               }
           }
           return C[n - 1];
       }
       #endregion

       #region 算法5
       private static string PPP_Hashcode(int a, int b)
       {
           return a.ToString() + "" + b.ToString();
       }

       public static int Min_Palindrome_Partiion_Memoizatoin(string input, int i, int j, Hashtable memo)
       {
           if (i > j)
           {
               return 0;
           }
           string ij = PPP_Hashcode(i, j);
           if (memo.ContainsKey(ij))
           {
               return (int)memo[ij];
           }

           if (i == j)
           {
               memo.Add(ij, 0);
               return 0;
           }
           if (PPP_IsPalindrome(input, i, j))
           {
               memo.Add(ij, 0);
               return 0;
           }
           int minimum = Int32.MaxValue;

           for (int k = i; k < j; k++)
           {
               int left_min = Int32.MaxValue;
               int right_min = Int32.MaxValue;
               string left = PPP_Hashcode(i, k);
               string right = PPP_Hashcode(k + 1, j);

               if (memo.ContainsKey(left))
               {
                   left_min = (int)memo[left];
               }
               if (memo.ContainsKey(right))
               {
                   right_min = (int)memo[right];
               }

               if (left_min == Int32.MaxValue)
               {
                   left_min = Min_Palindrome_Partiion_Memoizatoin(input, i, k, memo);
               }
               if (right_min == Int32.MaxValue)
               {
                   right_min = Min_Palindrome_Partiion_Memoizatoin(input, k + 1, j, memo);
               }
               minimum = Math.Min(minimum, left_min + 1 + right_min);
           }

memo.Add(ij, minimum);

return (int)memo[ij];
}
#endregion
}
}

POWER BY TRUFFER.CN

4 源代码

using System;
using System.Collections;
using System.Collections.Generic;namespace Legalsoft.Truffer.Algorithm
{public static partial class Algorithm_Gallery{#region 算法1private static bool PPP_IsPalindrome(string String, int i, int j){while (i < j){if (String[i] != String[j]){return false;}i++;j--;}return true;}public static int Min_Palindrome_Partion(string str, int i, int j){if (i >= j || PPP_IsPalindrome(str, i, j)){return 0;}int ans = Int32.MaxValue, count;for (int k = i; k < j; k++){count = Min_Palindrome_Partion(str, i, k) + Min_Palindrome_Partion(str, k + 1, j) + 1;ans = Math.Min(ans, count);}return ans;}#endregion#region 算法2public static int Min_Palindrome_Partion(string str){int n = str.Length;int[,] C = new int[n, n];bool[,] P = new bool[n, n];int i, j, k, L;for (i = 0; i < n; i++){P[i, i] = true;C[i, i] = 0;}for (L = 2; L <= n; L++){for (i = 0; i < n - L + 1; i++){j = i + L - 1;if (L == 2){P[i, j] = (str[i] == str[j]);}else{P[i, j] = (str[i] == str[j]) && P[i + 1, j - 1];}if (P[i, j] == true){C[i, j] = 0;}else{C[i, j] = int.MaxValue;for (k = i; k <= j - 1; k++){C[i, j] = Math.Min(C[i, j], C[i, k] + C[k + 1, j] + 1);}}}}return C[0, n - 1];}#endregion#region 算法3public static int Min_Cutting(string a){int[] cut = new int[a.Length];bool[,] palindrome = new bool[a.Length, a.Length];for (int i = 0; i < a.Length; i++){int minCut = i;for (int j = 0; j <= i; j++){if (a[i] == a[j] && (i - j < 2 || palindrome[j + 1, i - 1])){palindrome[j, i] = true;minCut = Math.Min(minCut, j == 0 ? 0 : (cut[j - 1] + 1));}}cut[i] = minCut;}return cut[a.Length - 1];}#endregion#region 算法4public static int Min_Palindrome_Partion_Second(string str){int n = str.Length;int[] C = new int[n];bool[,] P = new bool[n, n];int i, j, L;for (i = 0; i < n; i++){P[i, i] = true;}for (L = 2; L <= n; L++){for (i = 0; i < n - L + 1; i++){j = i + L - 1;if (L == 2){P[i, j] = (str[i] == str[j]);}else{P[i, j] = (str[i] == str[j]) && P[i + 1, j - 1];}}}for (i = 0; i < n; i++){if (P[0, i] == true){C[i] = 0;}else{C[i] = int.MaxValue;for (j = 0; j < i; j++){if (P[j + 1, i] == true && 1 + C[j] < C[i]){C[i] = 1 + C[j];}}}}return C[n - 1];}#endregion#region 算法5private static string PPP_Hashcode(int a, int b){return a.ToString() + "" + b.ToString();}public static int Min_Palindrome_Partiion_Memoizatoin(string input, int i, int j, Hashtable memo){if (i > j){return 0;}string ij = PPP_Hashcode(i, j);if (memo.ContainsKey(ij)){return (int)memo[ij];}if (i == j){memo.Add(ij, 0);return 0;}if (PPP_IsPalindrome(input, i, j)){memo.Add(ij, 0);return 0;}int minimum = Int32.MaxValue;for (int k = i; k < j; k++){int left_min = Int32.MaxValue;int right_min = Int32.MaxValue;string left = PPP_Hashcode(i, k);string right = PPP_Hashcode(k + 1, j);if (memo.ContainsKey(left)){left_min = (int)memo[left];}if (memo.ContainsKey(right)){right_min = (int)memo[right];}if (left_min == Int32.MaxValue){left_min = Min_Palindrome_Partiion_Memoizatoin(input, i, k, memo);}if (right_min == Int32.MaxValue){right_min = Min_Palindrome_Partiion_Memoizatoin(input, k + 1, j, memo);}minimum = Math.Min(minimum, left_min + 1 + right_min);}memo.Add(ij, minimum);return (int)memo[ij];}#endregion}
}