30. Substring with Concatenation of All Words

Posted 2020-11-16 wentiliangkaihua

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

 

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:
  s = "wordgoodgoodgoodbestword",
  words = ["word","good","best","word"]
Output: []

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> result = new ArrayList<>();
        if(s.length() == 0 || words.length == 0) return result;
        final int wordLength = words[0].length();
        final int catLength = wordLength * words.length;

        if (s.length() < catLength) return result;

        HashMap<String, Integer> wordCount = new HashMap<>();

        for (String word : words)
            wordCount.put(word, wordCount.getOrDefault(word, 0) + 1);

        for (int i = 0; i <= s.length() - catLength; ++i) {
            HashMap<String, Integer> unused = new HashMap<>(wordCount);

            for (int j = i; j < i + catLength; j += wordLength) {
                final String key = s.substring(j, j + wordLength);
                final int pos = unused.getOrDefault(key, -1);

                if (pos == -1 || pos == 0) break;

                unused.put(key, pos - 1);
                if (pos - 1 == 0) unused.remove(key);
            }

            if (unused.size() == 0) result.add(i);
        }

        return result;
    }
}

注意空s和空words的情况