Java 比较两个字符串并按字母顺序对它们进行排序
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Comparing two string and sorting them in alphabetical order
提问by
I want to compare two strings and sort them in alphabetical order. I am currently creating two arrays with the strings and sorting one them comparing the two arrays.
我想比较两个字符串并按字母顺序对它们进行排序。我目前正在用字符串创建两个数组，并比较两个数组对它们进行排序。
String a="LetterA";
String b="ALetterB";
String[] array1={a.toLowerCase(),b.toLowerCase()};
String[] array2={a.toLowerCase(),b.toLowerCase()};
Arrays.sort(array2);
if (Arrays.equals(array1, array2)){
System.out.println(a+" is before "+b);
}
else{
System.out.println(b+" is before "+a);
}
This works but it is time and memory consuming. I would appreciate if anyone can suggest a better way to do this.
这有效，但它是时间和内存消耗。如果有人能提出更好的方法来做到这一点，我将不胜感激。
采纳答案by Amit Sharma
Hint: All basic data type classes in java implement Comparable interface.
提示：java 中的所有基本数据类型类都实现了Comparable 接口。
String a="LetterA";
String b="ALetterB";
int compare = a.compareTo(b);
if (compare < 0){
System.out.println(a+" is before "+b);
}
else if (compare > 0) {
System.out.println(b+" is before "+a);
}
else {
System.out.println(b+" is same as "+a);
}
回答by irla
int compare = a.compareTo(b);
if (compare < 0){
System.out.println(a + " is before " +b);
} else if (compare > 0) {
System.out.println(b + " is before " +a);
} else {
System.out.println("Strings are equal")
}
回答by Anton Ivinskyi
If you just look for an easy and elegant code and you do not want to preoptimize, in java 8 you can do like this:
如果您只是寻找简单优雅的代码并且不想预先优化，那么在 java 8 中您可以这样做：
String[] sorted = Stream.of(a, b).sorted().toArray(String[]::new);
System.out.println(sorted[0] + " is before " + sorted[1]);
回答by Siva Sankaran
Here is the code for the problem.
这是问题的代码。
public static void main(String[] args) {
String a="Angram";
String b="Angram";
if(a.length()==b.length()) {
char c[]=a.toCharArray();
char d[]=b.toCharArray();
Arrays.sort(c);
Arrays.sort(d);
a=new String(c);
b=new String(d);
if(a.equals(b))
System.out.print("Yes");
else
System.out.print("No");
}
else {
System.out.print("No");
}
}
