真实业务算法

业务需求

1、公司相同，数字仅有一位不同，且有2个（含）以上相同数字连续且位置一致；

例如：CSN3473和CSN3438   CSN323和CSN325

2、公司相同，相同数字位置不完全一致；

例如：CDG4963和CDG4639

3、公司不同:

航班号是三位数字的后两位数字相同且位置一致；

航班号是四位数字的后三位数字相同且位置一致；

例如： CCA107和CSN307    CSN6555和CSC8555

       CSH9201和CSZ8201    

5、其他未涵盖但发音或外形比较容易混淆或易引起误听误判的呼号

    例如：OMA20144和OMA20044

我的解决办法

 

        ///     判断两个航班号是否相似
        /// </summary>
        /// <param name="calsign1"></param>
        /// <param name="calsign2"></param>
        /// <returns></returns>
        private static bool IsSameFlight(string calsign1, string calsign2)
        {
            int x = 0;//标记字母个数
            int y = 0;//标记数字个数

            //航班号位数不相同不是相似航班
            if (calsign1.Length != calsign2.Length) return false;

            //航班号位数相同
            for (int i = 0; i < calsign1.Length; i++)
            {
                
                if (calsign1[i] == calsign2[i] && (calsign1[i] >= 'A' && calsign1[i] <= 'Z'))
                {
                    x++;
                    continue;
                }

                #region 公司相同时，数字为3位的情况
                else if (x >= 3 && (calsign1.Length - i == 3))
                {
                    for (int j = i; j < calsign1.Length; j++)
                    {
                        if (calsign1[j] != calsign2[j] && y < 2)
                        {
                            if (!(calsign1[i] >= 'A' && calsign1[i] <= 'Z'))
                            {
                                int count = 0;
                                for (int k = i; k < calsign1.Length; k++)
                                    for (int l = i; l < calsign1.Length; l++)
                                        if (calsign1[k] == calsign2[l])
                                        {
                                            count++;
                                            break;
                                        }
                                if (count == 3) return true;
                            }
                            return false;
                        }
                        else if (calsign1[j] == calsign2[j] && y < 2)
                        {
                            y++;
                            continue;
                        }
                        else return true;
                    }


                } 
                #endregion

                #region 公司相同时，数字为4位的情况
                else if (x >= 3 && (calsign1.Length - i == 4))
                {
                    for (int j = i; j < calsign1.Length; j++)
                    {
                        if (calsign1[j] != calsign2[j] && y < 2)
                        {
                            if (!(calsign1[i] >= 'A' && calsign1[i] <= 'Z'))
                            {
                                int count = 0;
                                for (int k = i; k < calsign1.Length; k++)
                                    for (int l = i; l < calsign1.Length; l++)
                                        if (calsign1[k] == calsign2[l])
                                        {
                                            count++;
                                            break;
                                        }
                                if (count == 4) return true;
                            }
                            return false;
                        }
                        else if (calsign1[j] == calsign2[j] && y < 2)
                        {
                            y++;
                            continue;
                        }
                        else if (calsign1[j] != calsign2[j] && y == 2 && (calsign1[j] == calsign2[j + 1] || calsign1[j + 1] == calsign2[j] || calsign1[j + 1] == calsign2[j + 1]))
                            return true;
                        else return false;
                    }
                } 
                #endregion

                #region 公司不同时
                else if (x < 3)
                {
                    if (calsign1.Length == 6)
                        for (int j = calsign1.Length - 1; j > 2; j--)
                        {
                            if (calsign1[j] == calsign2[j])
                            {
                                y++;
                                continue;
                            }
                            else if (y < 2) return false;
                            else if (y >= 2) return true;
                        }
                    else if (calsign1.Length == 7)
                        for (int j = calsign1.Length - 1; j > 2; j--)
                        {
                            if (calsign1[j] == calsign2[j])
                            {
                                y++;
                                continue;
                            }
                            else if (y < 3) return false;
                            else if (y >= 3) return true;
                        }
                } 
                #endregion

            }

            //其他情况暂时为不相似
            return false;
        }

项目是针对具体应用的，所以看从业务逻辑到实现的过程比代码本身要重要。  其实不外乎选择、循环等结构。