如何在O（n）时间内检查两个字符串是否相互排列？ （Java）的

Question

我写了这个类，可以检查两个给定的字符串是否是彼此的排列。但是，据我所知，这是在O（n ^ 2）时间运行，因为string.indexOf（）在O（n）时运行。

如何提高这项计划的效率？

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;
   }
}

Answer 1

首先，可以通过对两个字符串进行排序（在将它们转换为O(nlogn)之后）在char[]中完成，然后简单的相等性测试将告诉您原始字符串是否是排列。 / p>

O(n)解决方案平均情况可以通过创建HashMap<Character, Integer>来实现，其中每个键都是字符串中的一个字符，值是它的出现次数（这称为{{{ 3}}）。拥有它之后，再次对两个地图进行简单的相等检查将告诉您原始字符串是否是排列。

Answer 2

存档O（n）的一种方法是计算每个字符的频率。

我会使用HashMap，其中字符为键，频率为值。

//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;
}

现在你可以为两个字符串创建一个HashMap，并比较每个字符的频率是否相同

//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;
}

编辑：（未经测试的）代码中存在nullpointer错误

Answer 3

排序解决方案：

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

  return Arrays.equals(sortedCharArray1, sortedCharArray2);
}

时间复杂度：O（n log n） 空间复杂度：O（n）

频率统计解决方案：

//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;
  }
}

时间复杂度：O（n） 空间复杂度：O（256）