字谜算法的java [英] Anagram algorithm in java
问题描述
我想作字谜算法,但 这code不起作用。哪里是我的错吗? 例如DES和sed是字谜,但输出并不字谜 同时,我不得不使用字符串的方法。不数组。 :)
公共静态布尔isAnagram(字符串S1,S2字符串)
{
字符串delStr =;
字符串中newstr =;
的for(int i = 0; I< s1.length();我++)
{
对于(INT J = 0; J< s2.length(); J ++)
{
如果(s1.charAt(I)== s2.charAt(J))
{
delStr = s1.substring(I,I + 1);
中newstr = s2.replace(delStr,);
}
}
}
如果(newStr.equals())
返回true;
其他
返回false;
}
这是更简单的方法可能是在两个字符串中的字符,只是排序,并比较它们是否相等:
公共静态布尔isAnagram(字符串S1,S2字符串){
//提前终止检查,如果字符串的长度不等,
//然后它们不能字谜
如果(s1.length()!= s2.length()){
返回false;
}
S1 = s1.toLowerCase();
S2 = s2.toLowerCase();
的char [] C1 = s1.toCharArray();
的char [] C2 = s2.toCharArray();
Arrays.sort(C1);
Arrays.sort(C2);
字符串SC1 =新的String(C1);
字符串SC2 =新的String(C2);
返回sc1.equals(SC2);
}
我个人认为这是更具可读性比嵌套的for循环= P
这有为O(n log n)的运行时的复杂性,其中 N
越长字符串的长度。
编辑:这不是最佳的解决方案。见@ aam1r的答案最有效的方法(例如,你其实应该说,在接受记者采访什么的)
I would like to make anagram algorithm but This code doesn't work. Where is my fault ? For example des and sed is anagram but output is not anagram Meanwhile I have to use string method. not array. :)
public static boolean isAnagram(String s1 , String s2)
{
String delStr="";
String newStr="";
for(int i=0;i<s1.length();i++)
{
for(int j=0 ; j < s2.length() ; j++)
{
if(s1.charAt(i)==s2.charAt(j))
{
delStr=s1.substring(i,i+1);
newStr=s2.replace(delStr,"");
}
}
}
if(newStr.equals(""))
return true;
else
return false;
}
An easier way might be to just sort the chars in both strings, and compare whether they are equal:
public static boolean isAnagram(String s1, String s2){
// Early termination check, if strings are of unequal lengths,
// then they cannot be anagrams
if ( s1.length() != s2.length() ) {
return false;
}
s1=s1.toLowerCase();
s2=s2.toLowerCase();
char[] c1 = s1.toCharArray();
char[] c2 = s2.toCharArray();
Arrays.sort(c1);
Arrays.sort(c2);
String sc1 = new String(c1);
String sc2 = new String(c2);
return sc1.equals(sc2);
}
Personally I think it's more readable than nested for-loops =p
This has O(n log n) run-time complexity, where n
is the length of the longer string.
Edit: this is not the optimal solution. See @aam1r's answer for the most efficient approach (i.e. what you should actually say in an interview)
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