从第二个字符串的第一个字符串中查找子字符串？

Question

更新：我有两个字符串。

  String a = "is am are has have was " //Helping verbs
  String b= " Robert and william are English Robert has a car he was living in England " //Sentence

更新：从字符串a中选择第一个子字符串，然后在字符串b的子字符串中搜索该子字符串。如果找到，则返回两者的索引号。然后继续在第二个String中找到第一个String的第二个subString，依此类推，直到last。

Answer 1

请在下面找到代码：-

        String stringA = "one two three four five six " ;
        String stringB = " eleven two four twelve thirteen ";


        String[] stringOne = stringA.split(" ");
        String[] stringTwo = stringB.trim().split(" ");

        Map<Integer,Integer> mapMatchingStrings = new HashMap<>();
        IntStream.range(0,stringOne.length).forEach(indexOfStringOne -> IntStream.range(0,stringTwo.length).forEach(indexOfStringTwo ->
                {
                 if(Objects.equals(stringOne[indexOfStringOne],stringTwo[indexOfStringTwo])){
                     mapMatchingStrings.put(indexOfStringOne,indexOfStringTwo);
                 }
                }
                ));
        System.out.println(mapMatchingStrings);

输出：-
{1 = 1，3 = 2}

Answer 2

拆分字符串的一种简单方法是使用StringTokenizer类

    String a = "one two three four five six";
    String b = "eleven two four twelve thirteen";
    StringTokenizer aT = new StringTokenizer(a," "/*delimiter(space or comma)*/);

    while (aT.hasMoreTokens()) {
        String indexA = aT.nextToken();
        StringTokenizer bT = new StringTokenizer(b," ");
        int index = 0;
        while(bT.hasMoreTokens()) {
            if(bT.nextToken().equals(indexA))System.out.println("Found Match:"+index);
            index++;
        }
    }

Answer 3

 String a = "one two three four five six ";
    String b= " eleven two four twelve thirteen "  ;
           List<String> aSub =   Arrays.asList(a.split(" "));
           List<String> bSub =   Arrays.asList(b.split(" "));

    for(String s1 :aSub){
        for(String s2 :bSub){
            if(s1.equalsIgnoreCase(s2)){
            System.out.println( s1+" exists and is found in index "+ bSub.indexOf(s1));
        }
      }
    }

结果：
存在两个并且在索引2中找到
存在四个并且在索引3中找到

Answer 4

//Big O of N*N solution
class SubString{
public static void main(String [] args)
{
System.out.println(new SubString().getCommonSubString("abcdefg","zycdemm"));
}
public int getCommonSubString(String A, String B)
{
int ans =0;

     for(int i=0;i<A.length();i++)
     {
       int c=0;
       int k =i;
       for(int j=0;k<A.length()&&j<B.length();j++)
       {

         if(A.charAt(k)==B.charAt(j))
         {
           c++;
           ans = Math.max(ans, c);
           k++;
         }else
         {
           c=0;
           k=i;
         }
      }
    }
 return ans;
}
}