串联所有单词的子串

/*
 * Problem: Given a string 's' and a list of words 'words', find all starting indices of substrings in 's' that are a concatenation of each word in 'words' exactly once and without any intervening characters.
 * 
 * Approach:
 * 1. Calculate the length of each word and the total length of the concatenated substring.
 * 2. Use a sliding window of the total length to traverse the string 's'.
 * 3. For each window, check if the substring can be broken into words in 'words'.
 * 4. Use Java Streams to check the frequency of each word in the window.
 */
 
import java.util.*;
import java.util.stream.*;
 
public class SolutionStream {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> result = new ArrayList<>();
        if (s == null || s.length() == 0 || words == null || words.length == 0) {
            return result;
        }
        int wordLength = words[0].length();
        int totalWords = words.length;
        int totalLength = wordLength * totalWords;
        Map<String, Long> wordCount = Arrays.stream(words)
                                             .collect(Collectors.groupingBy(w -> w, Collectors.counting()));
        for (int i = 0; i <= s.length() - totalLength; i++) {
            String current = s.substring(i, i + totalLength);
            List<String> parts = IntStream.range(0, totalWords)
                                           .mapToObj(j -> current.substring(j * wordLength, (j + 1) * wordLength))
                                           .collect(Collectors.toList());
            Map<String, Long> seenWords = parts.stream()
                                               .collect(Collectors.groupingBy(p -> p, Collectors.counting()));
            if (seenWords.equals(wordCount)) {
                result.add(i);
            }
        }
        return result;
    }
}