public static Boolean cmprStr( String s1, String s2 )
{
// STUFF
}
I want to iterate through s1 to make sure that ever开发者_开发知识库y character in s1 is included in s2.
public static Boolean cmprStr( String s1, String s2 )
{
for (int i = s1.length() - 1; i >= 0; --i) {
if (s2.indexOf(s1.charAt(i)) == -1) {
return Boolean.FALSE;
}
}
return Boolean.TRUE;
}
for(char c: s1.toCharArray()){
if(s2.indexOf(c) == -1){
return false;
}
}
return true;
Assuming that
s1 = "aabb";
s2 = "ccddaannbbss";
will return true.
length()
will give you the length of a string
charAt( someIndex)
will give you the character at a given position, so you can iterate the first String.
indexOf( achar )
will give you the poisition a char in a String, or -1 if it's not there. hence you should be able to look for each character in the first string within the second.
All the other answers are O(n^2). Here's a way that is linear in time (i.e. O(n)) using Google Guava:
public static boolean cmprStr(String s1, String s2) {
Set<Character> desiredCharacters = Sets.newHashSet(Lists.charactersOf(s2));
return Sets.difference(Sets.newHashSet(Lists.charactersOf(s1)), desiredCharacters).isEmpty();
}
Set<Character> charsInS1 = new HashSet<Character>();
for (int i = 0; i < s1.length(); i++) {
charsInS1.add(s1.charAt(i));
}
for (int i = 0; i < s2.length(); i++) {
charsInS1.remove(s2.charAt(i));
}
return charsInS1.isEmpty();
This has a complexity of O(n+m)
... answers using indexOf
have an O(n*m)
complexity. It does of course use a bit of extra memory temporarily though.
Why don't you simply use 'equals' method ?
Boolean b = s1.equals(s2);
Every String
is also a CharSequence
in Java. Therefore, you can easily iterate over a String
using a simple for loop:
int n = s.length();
for (int i = 0; i < n; ++i) {
char c = s.charAt(i);
...
}
As I understand the question it would be.
//for each character in s1
//if s2 does not contain character return false
//return true
for(int i = 0; i < length s1; i++){
if(!s2.contains(String.valueOf(s1.charAt(i)))){
return false;
}
}
return true;
This verifies that each character in s1 is in s2. It does not confirm order, nor how many are there, and is not an equals method.
Recursive:
public static Boolean cmprStr( String s1, String s2 )
{
if(s1.length() == 0 )
{
return true;
}
if(!s2.contains(s1.substring(0,1)))
{
return false;
}
return cmprStr(s1.substring(1), s2);
}
// Here's some code I wrote to find CG ratio in a gene
public double findCgRatio(String gene)
{
double cCount =0.0;
double gCount =0.0;
gene = gene.toLowerCase();
for(char character : gene.toCharArray())
{
if(character == 'c')
{
cCount++;
}
else if(character == 'g')
{
gCount++;
}
}
System.out.println("CG Ratio was :" + (cCount/gCount) );
return cCount/gCount; // cgRatio
}
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