在另一个字符串中找到一个字符串的 Anagrams 的最佳算法

Question

我在 Cracking The Coding Interview 一书中遇到了这个话题。挑战是在较大的字符串 b 中找到给定的较小字符串 s 的排列。我可以提出以下算法，其时间复杂度为 O(B x S)，其中 S 和 B 分别是给定较小和较大字符串的长度：

import java.util.HashMap;
public class AnagramAlgorithm {
public static void main(String[] args) {
    String s = "cbabadcbbabbcbabaabccbabc";
    String b = "abbc";

    printAnagramsOfB(s, b);
}

public static void printAnagramsOfB(String text, String pattern) {
    if(isEmpty(text) || isEmpty(pattern)) {
        System.out.println("Invalid Strings");
        return;
    }
    int patternLength = pattern.length();
    for (int i = 0; i < text.length() - patternLength + 1; i++) {
        String substring = text.substring(i, i + patternLength);
        if (isAnagram(pattern, substring)) {
            System.out.println("Anagram Found : " + substring);
        }
    }
}

public static boolean isEmpty(CharSequence str) {
    return str == null || str.length() == 0;
}

public static boolean isAnagram(String pattern, String substring) {
    if (pattern.length() != substring.length()) {
        System.out.println("SubString length doesn't match the length of Given String");
        return false;
    }
    char[] subStringArr = substring.toCharArray();
    char[] patternArr = pattern.toCharArray();
    HashMap<Character, Integer> mapPattern = new HashMap<>();
    HashMap<Character, Integer> mapSubstring = new HashMap<>();
    for (int i = 0; i < subStringArr.length; i++) {
        if (mapSubstring.containsKey(subStringArr[i])) {
            int count = mapSubstring.get(subStringArr[i]);
            mapSubstring.put(subStringArr[i], count + 1);
        } else {
            mapSubstring.put(subStringArr[i], 1);
        }
        if (mapPattern.containsKey(patternArr[i])) {
            int count = mapPattern.get(patternArr[i]);
            mapPattern.put(patternArr[i], count + 1);
        } else {
            mapPattern.put(patternArr[i], 1);
        }
    }
    return mapPattern.equals(mapSubstring);
}
}

书中提到最优算法是 O(B)。我想不出这样的算法。根据我的想法，对于整体复杂度为 O(B)，查找子字符串是否为字谜的算法应该是 O(1)，即没有任何循环。这甚至可能吗？或者有没有其他方法可以实现最优算法？

Answer 1

此算法以线性时间运行。如果您正在准备面试，那么您可能可以自己找出这里发生的事情;）

public class Solver {

    List<Integer> solve(String t, String s) {

        HashMap<Character, Integer> charCountInT = new HashMap<>();
        for (int i = 0; i < t.length(); i++) {
            Character c = t.charAt(i);
            if (charCountInT.containsKey(c)) {
                charCountInT.put(c, charCountInT.get(c) + 1);
            }
            else {
                charCountInT.put(c, 1);
            }
        }

        HashMap<Character, Integer> extraCharacters = new HashMap<>();
        for (Character c : charCountInT.keySet()) {
            extraCharacters.put(c, -charCountInT.get(c));
        }
        for (int i = 0; i < t.length(); i++) {
            Character c = s.charAt(i);
            if (extraCharacters.containsKey(c)) {
                extraCharacters.put(c, extraCharacters.get(c) + 1);
            }
        }

        int expectedZeroesInExtraCharacters = charCountInT.size();
        int zeroesInExtraCharacters = 0;
        for (Integer count : extraCharacters.values()) {
            if (count == 0) ++zeroesInExtraCharacters;
        }

        List<Integer> answer = new ArrayList<>();
        if (zeroesInExtraCharacters == expectedZeroesInExtraCharacters) answer.add(0);

        for (int i = 1; i < s.length() - t.length(); i++) {

            Character nextChar = s.charAt(t.length() + i - 1);
            if (charCountInT.containsKey(nextChar)) {
                extraCharacters.put(nextChar, extraCharacters.get(nextChar) + 1);
                if (extraCharacters.get(nextChar) == 0) ++zeroesInExtraCharacters;
                if (extraCharacters.get(nextChar) == 1) --zeroesInExtraCharacters;
            }

            Character removedChar = s.charAt(i - 1);
            if (charCountInT.containsKey(removedChar)) {
                extraCharacters.put(removedChar, extraCharacters.get(removedChar) - 1);
                if (extraCharacters.get(removedChar) == 0) ++zeroesInExtraCharacters;
                if (extraCharacters.get(removedChar) == -1) --zeroesInExtraCharacters;
            }

            if (zeroesInExtraCharacters == expectedZeroesInExtraCharacters) answer.add(i);
        }

        return answer;

    }

    public static void main(String[] args) {
        String t = "abbc";
        String s = "cbabadcbbabbcbabaabccbabc";
        List<Integer> startIndices = new Solver().solve(t, s);
        System.out.println(startIndices);
        for (int startIndex : startIndices) {
            System.out.println(s.substring(startIndex, startIndex + t.length()));
        }
    }

}

Answer 2

标准字谜查找方法使用排序版本进行比较。您不寻找“TARGET”，而是先将其按字母顺序排列：“AEGRTT”并搜索。

从较大字符串的开头获取相关的字符数，“AEGRTT”为六个。排序并比较以查看是否匹配。然后删除第一个字母并将长列表中的下一个字母添加到适当的位置 - 列表已经排序，这将有所帮助。再次比较并重复。