如何检查两个字符串是否在 O(n) 时间内相互排列？ （爪哇）

Question

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

如何让这个程序更有效率？

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，可以在O(nlogn)中对两个字符串进行排序（将它们转换为char[]之后），然后简单的相等测试会告诉你原始字符串是否是排列。

可以通过创建HashMap<Character, Integer> 来实现O(n) 解决方案平均情况，其中每个键是字符串中的一个字符，值是它的出现次数（这称为Histogram）。获得它之后，再次对两个映射进行简单的相等性检查将告诉您原始字符串是否是排列。

Answer 2

归档 O(n) 的一种方法是计算每个字符出现的频率。

我会使用 HashMap，其中字符作为键，频率作为值。

//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;
}

现在您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）代码中存在空指针错误

Answer 3

分拣解决方案：

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)

频率计数解决方案：

//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)