两个字符串的最长相同子序列

描述：

      求解两个字符串的最长子序列。输出其长度和子字符串

输入：

       S1:abchifkdacb

       S2:bcacbhiab

输出：

           bchiab

解题思路：

       可以利用求解出两个子字符串的最长长度，再求出其完成字符串的最长长度。

        S1[0...i]与S2[0...j]的长度等于：{S1[0..i-1]与S2[0..j-1]加1}(此时S1[i]==S2[j])

                                                       {S1[0..i-1]与S2[0..j]和S1[0..i]与S2[0..j-1]中长度较大的那个}

求解子序列：

/// <summary>
        /// 
        /// </summary>
        /// <param name="firtLength"></param>
        /// <param name="secondLength"></param>
        /// <param name="S1">第一个字符串</param>
        /// <param name="S2">第二个字符串</param>
        /// <param name="M">计算长度数组</param>
        /// <param name="N">是否相同数组</param>
        public static void LCSLength(int firtLength, int secondLength, string S1, string S2, ref int[,] M, ref int[,] N)
        {
            if (firtLength == 0 || secondLength == 0)
                return;
            //从第一列开始向右逐个求解，求出最长子序列
            for (int i = 1; i < firtLength; i++)
                for (int j = 1; j < secondLength; j++)
                {
                    if (S1[i] == S2[j])
                    {
                        N[i, j] = 1;
                        M[i, j] = M[i - 1, j - 1] + 1;
                    }
                    else
                    {
                        if (M[i - 1, j] > M[i, j - 1])
                        {
                            M[i, j] = M[i - 1, j];
                            N[i, j] = 2;
                        }
                        else
                        {
                            M[i, j] = M[i, j - 1];
                            N[i, j] = 3;
                        }
                    }
                }
        }

输出子序列：

/// <summary>
        /// 输出其子字符串
        /// </summary>
        /// <param name="firstLength"></param>
        /// <param name="secondLength"></param>
        /// <param name="S1"></param>
        /// <param name="N"></param>
        public static void LCS(int firstLength, int secondLength, string S1, int[,] N)
        {
            if (firstLength == 0 || secondLength == 0)
                return;
            if (N[firstLength, secondLength] == 1)
            {
                LCS(firstLength - 1, secondLength - 1, S1, N);
                Console.Write(S1[firstLength]);
            }
            else
            {
                if (N[firstLength, secondLength] == 2)
                {
                    LCS(firstLength - 1, secondLength, S1, N);
                }
                if (N[firstLength, secondLength] == 3)
                {
                    LCS(firstLength, secondLength - 1, S1, N);
                }
            }
        }