Java e。（或int[128]表示ASCII）+1表示文本，通常称为频率计数。“直方图”更常用于数字的分布。（字符是我知道的数字，但通常不这么认为）可以使用int[65536]而不是HashMap来持续使用内存。（或int[128]表示ASCII）+1表示文本，

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e。（或

int[128]

表示ASCII）+1表示文本，通常称为频率计数。“直方图”更常用于数字的分布。（字符是我知道的数字，但通常不这么认为）可以使用

int[65536]

而不是HashMap来持续使用内存。（或

int[128]

表示ASCII）+1表示文本，通常称为频率计数。“直方图”更常用于数字的分布。（字符是我知道的数字，但人们通常不这么认为）

import java.util.*;

public class IsPermutation{
   public void IsPermutation(){
      System.out.println("Checks if two strings are permutations of each other.");
      System.out.println("Call the check() method");
   }

   public boolean check(){
      Scanner console = new Scanner(System.in);
      System.out.print("Insert first string: ");
      String first = console.nextLine();
      System.out.print("Insert second string: ");
      String second = console.nextLine();

      if (first.length() != second.length()){
         System.out.println("Not permutations");
         return false;
      }

      for (int i = 0; i < first.length(); i++){
         if (second.indexOf(first.charAt(i)) == -1){
            System.out.println("Not permutations");
            return false;
         } 
      }
      System.out.println("permutations");
      return true;
   }
}

//create a HashMap containing the frequencys of every character of the String  (runtime O(n) )
public HashMap<Character, Integer> getFrequencys(String s){
    HashMap<Character, Integer> map = new HashMap<>();

    for(int i=0; i<s.length(); i++){
        //get character at position i
        char c = s.charAt(i);

        //get old frequency (edited: if the character is added for the 
        //first time, the old frequency is 0)
        int frequency;
        if(map.containsKey(c)){
            frequency = map.get(c);
        }else{
            frequency = 0;
        }
        //increment frequency by 1
        map.put(c, frequency+1 );
    }

    return map;
}

//runtime O(3*n) = O(n)
public boolean compare(String s1, String s2){
    if(s1.length() != s2.length()){
        return false;
    }

    //runtime O(n)
    HashMap<Character, Integer> map1 = getFrequencys(s1);
    HashMap<Character, Integer> map2 = getFrequencys(s2);

    //Iterate over every character in map1 (every character contained in s1)  (runtime O(n) )
    for(Character c : map1.keySet()){
        //if the characters frequencys are different, the strings arent permutations
        if( map2.get(c) != map1.get(c)){
            return false;
        }
    }

    //since every character in s1 has the same frequency in s2,
    //and the number of characters is equal => s2 must be a permutation of s1

    return true;
}

public void IsPermutation(String str1, String str2) {
  char[] sortedCharArray1 = Arrays.sort(str1.toCharArray());
  char[] sortedCharArray2 = Arrays.sort(str2.toCharArray());

  return Arrays.equals(sortedCharArray1, sortedCharArray2);
}

//Assuming that characters are only ASCII. The solutions can easily be modified for all characters

public void IsPermutation(String str1, String str2) {
    if (str1.length() != str2.length())
        return false;

    int freqCountStr1[] = new int[256];
    int freqCountStr2[] = new int[256];

    for (int i = 0; i < str1.length(); ++i) {
        int c1 = str1.charAt(i);
        int c2 = str2.charAt(i);
        ++freqCountStr1[c1];
        ++freqCountStr2[c2];
    }

    for (int i = 0; i < str1.length(); ++i) {
        if (freqCountStr1[i] != freqCountStr2[i]) {
            return false;
        }
    }

    return true;
  }
}