java-字符串处理

luxuantao

发布于 2021-02-24 11:23:55

3650

发布于 2021-02-24 11:23:55

java的字符串处理，有涉及到HushMap和其他一些零散知识点的应用，作为初学者，这题就作为范例供来日所需。

Two strings, and , are called anagrams if they contain all the same characters in the same frequencies. For example, the anagrams of CAT are CAT, ACT, TAC, TCA, ATC, and CTA.

Complete the function in the editor. If and are case-insensitive anagrams, print “Anagrams”; otherwise, print “Not Anagrams” instead.

Input Format

The first line contains a string denoting . The second line contains a string denoting .

Constraints

Strings and consist of English alphabetic characters.
The comparison should NOT be case sensitive.

Output Format

Print “Anagrams” if and are case-insensitive anagrams of each other; otherwise, print “Not Anagrams” instead.

Sample Input 0

anagram
margana

Sample Output 0

Anagrams

Explanation 0

Character	Frequency: anagram	Frequency: margana
A or a	3	3
G or g	1	1
N or n	1	1
M or m	1	1
R or r	1	1

The two strings contain all the same letters in the same frequencies, so we print “Anagrams”.

Sample Input 1

anagramm
marganaa

Sample Output 1

Not Anagrams

Explanation 1

Character	Frequency: anagramm	Frequency: marganaa
A or a	3	4
G or g	1	1
N or n	1	1
M or m	2	1
R or r	1	1

The two strings don’t contain the same number of a‘s and m‘s, so we print “Not Anagrams”.

Sample Input 2

Hello
hello

Sample Output 2

Anagrams

Explanation 2

Character	Frequency: Hello	Frequency: hello
E or e	1	1
H or h	1	1
L or l	2	2
O or o	1	1

The two strings contain all the same letters in the same frequencies, so we print “Anagrams”.

代码如下：

import java.io.*;
import java.util.*;
public class Solution {
  static boolean isAnagram(String a, String b) {
        if (a.length()!=b.length())
            return false;
        Map<Character, Integer> map = new HashMap<>();
        a=a.toLowerCase();
        b=b.toLowerCase();
        for(int i=0;i<a.length();i++) {
            char ch=a.charAt(i);
            if(!map.containsKey(ch)) {
                map.put(ch,1);
            } 
            else {
                Integer frequency = map.get(ch);
                map.put(ch,++frequency);
            }
        }
        for(int i=0;i<b.length();i++) {
            char ch=b.charAt(i);
            if(!map.containsKey(ch)) {
                return false;
            } 
            Integer frequency = map.get(ch);
            if(frequency==0) {
                return false;
            }
            else {
                map.put(ch,--frequency);
            }
        }
        return true;
  }
  public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String a = scan.next();
        String b = scan.next();
        scan.close();
        boolean ret = isAnagram(a, b);
        System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
  }
}

方法二：

import java.util.*;

public class Solution {

    static boolean isAnagram(String s1, String s2) {
        if (s1.length() != s2.length()) {
            return false;
        }
        
        char[] a = s1.toLowerCase().toCharArray();
        char[] b = s2.toLowerCase().toCharArray();
        boolean anagram = true;
        Arrays.sort(a);
        Arrays.sort(b);
        
        for(int i = 0; i < a.length; i++) {
            if(a[i] != b[i]) {
                anagram = false;
            }
        }
        
        return anagram;
    }
  
    public static void main(String[] args) {
    
        Scanner scan = new Scanner(System.in);
        String a = scan.next();
        String b = scan.next();
        scan.close();
        boolean ret = isAnagram(a, b);
        System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
    }
}

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原始发表：2017-09-15，如有侵权请联系 cloudcommunity@tencent.com 删除

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