varadhan1127
BAN USERimport java.util.*;
class samp
{
public static void main(String args[])
{
String s="abc";
String s1;
StringBuilder sb= new StringBuilder();
int i,j=0;
int a=s.length();
for(i=0;i<=a;i++)
{
while(j<=a)
{
s1=s.substring(i,j);
sb.append(s1);
j++;
}
j=i+1;
}
System.out.println(sb);
}
}
import java.util.Scanner;
class samp
{
private static Scanner scan;
private String mini(String s1,String[]s2)
{
StringBuilder bu = new StringBuilder();
String sb[]=null;
int k=s1.length();
while(k>0 )
{
for(int i=0;i<s2.length;i++)
{
sb=s1.split(s2[i], s2[i].length());
for(int j=0;j<sb.length;j++)
{
s1=sb[j];
bu.append(s1);
s1=bu.toString();
}
bu.setLength(0);
}
k--;
}
return s1;
}
public static void main(String args[])
{
String str="ccdaabcdbb";
String[] sstr={"ab","cd"};
String nstr=null;
samp t= new samp();
nstr=t.mini(str,sstr);
System.out.println("String is: "+nstr+"\n"+"Length is: "+nstr.length());
}
}
We can implement using below graph by getting the adjacent value then check for increasing order needs extra effort.
- varadhan1127 July 14, 20141 2 3 4 5 6 7 8 9
1 0 1 0 0 1 0 0 0 0
2 1 0 1 1 1 0 0 0 0
3 0 1 0 0 1 1 0 1 0
4 0 1 0 0 0 1 0 0 0
5 1 0 1 0 0 0 0 0 1
6 0 0 1 1 0 0 1 0 0
7 0 0 0 0 0 1 0 1 0
8 0 0 1 0 0 0 1 0 1
9 0 0 0 0 1 0 0 1 0
the sequence can be pushed into the stack for the result.
Suggest if the above solution is wrong.