字符串相似性判断？

Question

string s1="8405";
string s2="8425";if(s1.ToList()..Intersect(s2ToList().).ToList().Count()>=3)//交集
{
   //相似
}
else
{
//不相似
}

Answer 1

好像不对，比如8998,8908判断是false,实际情况是true

Answer 2

            string s1 = this.textBox1.Text.Trim();
            string s2 = this.textBox2.Text.Trim();
            int count = 0;
            foreach (char c1 in s1.ToArray())
            {
                foreach (char c2 in s2.ToArray())
                {
                    if (c1 == c2)
                    {
                        count++;
                        break;
                    }
                }
            }
            if (count >= 3)
            {
                Console.WriteLine("相似");
            }
            else
            {
                Console.WriteLine("不相似");
            }

Answer 3

7334,3775，按照你的方法为true,实际情况为false,也不对

Answer 4

应该先对字符串进行排序
然后分别循环,截取子串,3个字符,对比,判断contains

Answer 5

方法比较笨，先各自排序，然后找连续三个字符是否相等
   string s1 = "8990";
            string s2 = "9989";
            char[] arry1 = s1.ToArray().OrderBy(x => x).ToArray();  //排序
            char[] arry2 = s2.ToArray().OrderBy(x => x).ToArray();  //排序
            int count = 0;
            for (int i = 0; i < arry1.Length; i++)
            {
                for (int j = 0; j < arry2.Length; j++)
                {
                    if (arry1[i] == arry2[j]) //有一个相等，判断后面2个是否也相等，相等则认为相似
                    {
                        count = 1;
                        bool flag = false;
                        for (int x = 0; x < 3; x++) //判断后面2个是否也相等
                        {
                            if (i + x + 1 >= arry1.Length || j + x + 1 >= arry2.Length)
                                break;                            if (arry1[i + x + 1] != arry2[j + x + 1])  //不相等，则此连续段不相似
                            {
                                flag = true;
                                break;
                            }
                            count++;                        }
                        if (flag) //此连续段不相似，重新往后判断
                            break;
                        if (count >= 3)
                        {
                            MessageBox.Show("相似");
                            return;
                        }
                    }
                }
            }
            if (count < 3)
            {
                MessageBox.Show("不相似");
            }

Answer 6

string s1 = "1234";
string s2 = "2345";
bool 相似 = s1.Join(s2, x => x, x => x, (x, y) => x == y).Count(x => x) > 3;

Answer 7

7334,3775，按照你的方法为true,实际情况为false,也不对
他按照你的要求 他写的就是对的。。如果不对请把 c1 == c2改成值比较c1.Equals(c2)

Answer 8

 public class StringMatch
    {
        public static int Match(string s1,string s2)
        {
           //按最短的字符串查找
            if (s1.Length>s2.Length)
            {
                string stmp = s1;
                s1 = s2;
                s2 = s1;
            }           int iMatch = 0;
           for (int i = 0; i < s1.Length; i++)
           {
               int c1 = GetCharCount(s1, s1.Substring(i, 1),i);
               int c2 = GetCharCount(s2, s1.Substring(i, 1));
               if (c1<=c2)
               {
                   iMatch++;
               }
           }
           return iMatch;
        }
        public static int GetCharCount(string str, string sChar)
        {
            int pos = 0;
            int Count = 0;
            while (pos != -1)
            {
                pos = str.IndexOf(sChar, pos);//获取字符的索引
                if (pos != -1)
                {
                    Count++;
                    pos++;                }
            }
            return Count;        }
        public static int GetCharCount(string str,string sChar,int endPos)
        {
            int pos = 0;
            int Count = 0;
            while (pos != -1)
            {
                pos = str.IndexOf(sChar, pos);//获取字符的索引
                if (pos != -1)
                {
                    if (pos>endPos)
                    {
                        return Count;
                    }
                    else
                    {
                        Count++;
                        pos++;
                    }
                }
            }
            return Count;        }
    }
测试结果： int m=  StringMatch.Match("1234", "1235");

Answer 9

http://www.cnblogs.com/liangxiaxu/archive/2012/05/05/2484972.html

Answer 10

nnd，试了下面几组都是可以了，
楼主再举例再不行，我也没招了
string s1 = "8998"; //"8998","7334","1245"
            string s2 = "8908"; //"8908","3775","1345"            var ex = s1.ToArray().Intersect(s2.ToArray()).ToArray(); //交集
            int count = 0;
            foreach (char c in ex) //看交集在s1和s2中出现的次数
            {
                int x1 = 0, x2 = 0;
                foreach (char c1 in s1.ToArray())
                    if (c == c1)
                        x1++;
                foreach (char c2 in s2.ToArray())
                    if (c == c2)
                        x2++;                count += Math.Min(x1, x2); //取最小的次数
            }
            if (count >= 3)
            {
                MessageBox.Show("相似");
            }
            else
            {
                MessageBox.Show("不相似哦");
            }

Answer 11

http://my.oschina.net/BreathL/blog/42477

Answer 12

字符串a，字符串b,将字符串a，变量i，的每个字符存入List<string>() ,再循环遍历字符b的每个字符在List<string>()中是否存在，存在则i+1，便利完毕判断存在数是否 >=3
/// <summary>
    /// 验证字符StrB和StrA是否相似
    /// </summary>
    /// <param name="StrA"></param>
    /// <param name="StrB"></param>
    /// <returns></returns>
    public bool IsLike(string StrA, string StrB)
    {
        List<string> list = new List<string>();
        int count = 0;
        for (int i = 0; i < StrA.Length; i++)
        {
            string str = StrA.Substring(i, 1);
            list.Add(str);
        }
        for (int i = 0; i < StrB.Length; i++)
        {
            string str = StrB.Substring(i, 1);
            if (list.Contains(str))
                count++;
        }
        if (count >= 3)
            return true;
        else
            return false;
    }

Answer 13

有何难,把之前写的一个稍改即可
            string aa1 = "9898", aa2 = "8908", aa3 = "";
            for (int a = 0; a < aa1.Length - 1; a++)
                if (aa2.Contains(aa1.Substring(a, 1)))
                    aa3 += aa1.Substring(a, 1);http://blog.csdn.net/xianfajushi/article/details/11741799

Answer 14

随便写个            var aa = "9898,8908".Split(',');/*字符所有相似计算*/
            var bb = aa[0].Select(cc => aa[1].Contains(cc.ToString())).Count(ee => ee == true);
            var dd = aa[1].Select(cc => aa[0].Contains(cc.ToString())).Count(ee => ee == true);
if(bb>=3&&dd>=3)...

Answer 15

对这些函数不是很熟悉多练习            var ff = aa[1].Where(gg => aa[0].Contains(gg)).Count();
            var hh = aa[0].Where(gg => aa[1].Contains(gg)).Count();

Answer 16

static void Main(string[] args)
        {
            string str1 = "8990";
            string str2 = "99089";
            Hashtable ht1 = getHt(str1);
            Hashtable ht2 = getHt(str2);            int iCount = 0;
            foreach (char item1 in ht1.Keys)
            {
                foreach (char item2 in ht2.Keys)
                {
                    if (item1 == item2)
                    {
                        iCount++;
                        break;
                    }
                }
            }            if (iCount > 2)
            {                Console.WriteLine("相似");
            }
            else
            {
                Console.WriteLine("不相似");
            }
            Console.ReadKey();
        }        public static  Hashtable getHt(string s)
        {
            Hashtable ht = new Hashtable();
            foreach (char item in s)
            {
                if (ht.Contains(item))
                {
                    int icount = (int)ht[item];
                    icount++;
                    ht[item] = icount;                }
                else
                {
                    ht.Add(item, 1);
                }
            }            return
                ht;
        }