Question

在这里开始CS学生，试图掌握循环。我的任务是获取一个String，这样如果s = 'abcd'，程序将打印：'a,b,c,d,ab,bc,cd,abc,bcd,abcd'。显然，begin索引和end索引随循环迭代而变化 - (0,1),(1,2),(2,3),(3,4)。那很好，我可以打印：'a,b,c,d'。但是我如何控制它，使其从(0,2),(1,3),(2,4)再到(0,3),(1,4),，最后(0,4)？这是我难倒的地方。感谢您的帮助，这是我的代码：

void printSubstring()
{
    int len = s.length();

    for (int i=0;i<len;i++)
    {
        for (int k=0;k<len;k++)
           System.out.print(s.substring(k,k+1)+", "); 
    }
    System.out.println();
}

Answer 1

for (int i=0;i<len;i++)
{
    for (int k=0;k<len-i;k++)
       System.out.print(s.substring(k,k+i+1)+","); 
}

这对你的结构应该有效。

在每个步骤内循环中，要打印的字符串数量减少（使用-i）。并且字符串的长度增加（在子字符串中使用+ i）

经过测试：abcdefghijklmnopqrstuvwxyz

输出：

a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,ab,bc,cd,de,ef,fg,gh,hi,ij,jk,kl,lm,mn,no,op,pq,qr,rs,st,tu,uv,vw,wx,xy,yz,abc,bcd,cde,def,efg,fgh,ghi,hij,ijk,jkl,klm,lmn,mno,nop,opq,pqr,qrs,rst,stu,tuv,uvw,vwx,wxy,xyz,abcd,bcde,cdef,defg,efgh,fghi,ghij,hijk,ijkl,jklm,klmn,lmno,mnop,nopq,opqr,pqrs,qrst,rstu,stuv,tuvw,uvwx,vwxy,wxyz,abcde,bcdef,cdefg,defgh,efghi,fghij,ghijk,hijkl,ijklm,jklmn,klmno,lmnop,mnopq,nopqr,opqrs,pqrst,qrstu,rstuv,stuvw,tuvwx,uvwxy,vwxyz,abcdef,bcdefg,cdefgh,defghi,efghij,fghijk,ghijkl,hijklm,ijklmn,jklmno,klmnop,lmnopq,mnopqr,nopqrs,opqrst,pqrstu,qrstuv,rstuvw,stuvwx,tuvwxy,uvwxyz,abcdefg,bcdefgh,cdefghi,defghij,efghijk,fghijkl,ghijklm,hijklmn,ijklmno,jklmnop,klmnopq,lmnopqr,mnopqrs,nopqrst,opqrstu,pqrstuv,qrstuvw,rstuvwx,stuvwxy,tuvwxyz,abcdefgh,bcdefghi,cdefghij,defghijk,efghijkl,fghijklm,ghijklmn,hijklmno,ijklmnop,jklmnopq,klmnopqr,lmnopqrs,mnopqrst,nopqrstu,opqrstuv,pqrstuvw,qrstuvwx,rstuvwxy,stuvwxyz,abcdefghi,bcdefghij,cdefghijk,defghijkl,efghijklm,fghijklmn,ghijklmno,hijklmnop,ijklmnopq,jklmnopqr,klmnopqrs,lmnopqrst,mnopqrstu,nopqrstuv,opqrstuvw,pqrstuvwx,qrstuvwxy,rstuvwxyz,abcdefghij,bcdefghijk,cdefghijkl,defghijklm,efghijklmn,fghijklmno,ghijklmnop,hijklmnopq,ijklmnopqr,jklmnopqrs,klmnopqrst,lmnopqrstu,mnopqrstuv,nopqrstuvw,opqrstuvwx,pqrstuvwxy,qrstuvwxyz,abcdefghijk,bcdefghijkl,cdefghijklm,defghijklmn,efghijklmno,fghijklmnop,ghijklmnopq,hijklmnopqr,ijklmnopqrs,jklmnopqrst,klmnopqrstu,lmnopqrstuv,mnopqrstuvw,nopqrstuvwx,opqrstuvwxy,pqrstuvwxyz,abcdefghijkl,bcdefghijklm,cdefghijklmn,defghijklmno,efghijklmnop,fghijklmnopq,ghijklmnopqr,hijklmnopqrs,ijklmnopqrst,jklmnopqrstu,klmnopqrstuv,lmnopqrstuvw,mnopqrstuvwx,nopqrstuvwxy,opqrstuvwxyz,abcdefghijklm,bcdefghijklmn,cdefghijklmno,defghijklmnop,efghijklmnopq,fghijklmnopqr,ghijklmnopqrs,hijklmnopqrst,ijklmnopqrstu,jklmnopqrstuv,klmnopqrstuvw,lmnopqrstuvwx,mnopqrstuvwxy,nopqrstuvwxyz,abcdefghijklmn,bcdefghijklmno,cdefghijklmnop,defghijklmnopq,efghijklmnopqr,fghijklmnopqrs,ghijklmnopqrst,hijklmnopqrstu,ijklmnopqrstuv,jklmnopqrstuvw,klmnopqrstuvwx,lmnopqrstuvwxy,mnopqrstuvwxyz,abcdefghijklmno,bcdefghijklmnop,cdefghijklmnopq,defghijklmnopqr,efghijklmnopqrs,fghijklmnopqrst,ghijklmnopqrstu,hijklmnopqrstuv,ijklmnopqrstuvw,jklmnopqrstuvwx,klmnopqrstuvwxy,lmnopqrstuvwxyz,abcdefghijklmnop,bcdefghijklmnopq,cdefghijklmnopqr,defghijklmnopqrs,efghijklmnopqrst,fghijklmnopqrstu,ghijklmnopqrstuv,hijklmnopqrstuvw,ijklmnopqrstuvwx,jklmnopqrstuvwxy,klmnopqrstuvwxyz,abcdefghijklmnopq,bcdefghijklmnopqr,cdefghijklmnopqrs,defghijklmnopqrst,efghijklmnopqrstu,fghijklmnopqrstuv,ghijklmnopqrstuvw,hijklmnopqrstuvwx,ijklmnopqrstuvwxy,jklmnopqrstuvwxyz,abcdefghijklmnopqr,bcdefghijklmnopqrs,cdefghijklmnopqrst,defghijklmnopqrstu,efghijklmnopqrstuv,fghijklmnopqrstuvw,ghijklmnopqrstuvwx,hijklmnopqrstuvwxy,ijklmnopqrstuvwxyz,abcdefghijklmnopqrs,bcdefghijklmnopqrst,cdefghijklmnopqrstu,defghijklmnopqrstuv,efghijklmnopqrstuvw,fghijklmnopqrstuvwx,ghijklmnopqrstuvwxy,hijklmnopqrstuvwxyz,abcdefghijklmnopqrst,bcdefghijklmnopqrstu,cdefghijklmnopqrstuv,defghijklmnopqrstuvw,efghijklmnopqrstuvwx,fghijklmnopqrstuvwxy,ghijklmnopqrstuvwxyz,abcdefghijklmnopqrstu,bcdefghijklmnopqrstuv,cdefghijklmnopqrstuvw,defghijklmnopqrstuvwx,efghijklmnopqrstuvwxy,fghijklmnopqrstuvwxyz,abcdefghijklmnopqrstuv,bcdefghijklmnopqrstuvw,cdefghijklmnopqrstuvwx,defghijklmnopqrstuvwxy,efghijklmnopqrstuvwxyz,abcdefghijklmnopqrstuvw,bcdefghijklmnopqrstuvwx,cdefghijklmnopqrstuvwxy,defghijklmnopqrstuvwxyz,abcdefghijklmnopqrstuvwx,bcdefghijklmnopqrstuvwxy,cdefghijklmnopqrstuvwxyz,abcdefghijklmnopqrstuvwxy,bcdefghijklmnopqrstuvwxyz,abcdefghijklmnopqrstuvwxyz,

Answer 2

内循环应该是 -

for (int k=0;k<len-i;k++)
       System.out.print(s.substring(k,k+i+1)+", ");

希望有所帮助

Answer 3

void printSubstring() {

    int len = s.length();

    for (int i=1;i<len+1;i++)
    {
        for (int k=0;k<len-i;k++)
        System.out.print(s.substring(k,k+i)+", "); 
    }
    System.out.println();
}

Answer 4

你可以尝试这样的事情，

void printSubstring(String s){
  if(s == null){
    return;
  }

  for(int lenghtSubstring = 1; lenghtSubstring <= s.length(); lenghtSubstring++){
    for(int index = 0; index <= s.length() - lenghtSubstring; index++){
      System.out.print(s.substring(index, index + lenghtSubstring) + ",");
    }
  }
}

用于带子串的循环

4 个答案: