30. Substring with Concatenation of All Words

难度: Hard

题目描述

You are given a string s and an array of strings words. All the strings of words are of the same length.

A concatenated string is a string that exactly contains all the strings of any permutation of words concatenated.

• For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated string because it is not the concatenation of any permutation of words.

Return an array of the starting indices of all the concatenated substrings in s. You can return the answer in any order.

 

Example 1:

Input: s = "barfoothefoobarman", words = ["foo","bar"]

Output: [0,9]

Explanation:

The substring starting at 0 is "barfoo". It is the concatenation of ["bar","foo"] which is a permutation of words.
The substring starting at 9 is "foobar". It is the concatenation of ["foo","bar"] which is a permutation of words.

Example 2:

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

Output: []

Explanation:

There is no concatenated substring.

Example 3:

Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]

Output: [6,9,12]

Explanation:

The substring starting at 6 is "foobarthe". It is the concatenation of ["foo","bar","the"].
The substring starting at 9 is "barthefoo". It is the concatenation of ["bar","the","foo"].
The substring starting at 12 is "thefoobar". It is the concatenation of ["the","foo","bar"].

 

Constraints:

• 1 <= s.length <= 104
• 1 <= words.length <= 5000
• 1 <= words[i].length <= 30
• s and words[i] consist of lowercase English letters.

解题思路

代码实现

解决方案

java

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {

        List<Integer> res = new ArrayList<>();

        int sLen = s.length();
        int singleWordLen = words[0].length();
        int wordsLen = words.length;

        // i>singleWordLen 的情况 会被 i< singleWordLen 重复 所以最大界限是 singleWordLen
        for (int i = 0; i < singleWordLen; i++) {

            if (i + singleWordLen * wordsLen > sLen) {
                break;
            }

            Map<String, Integer> differ = new HashMap<>();

            for (int j = 0; j < wordsLen; j++) {
                String word = s.substring(i + j * singleWordLen, i + (j + 1) * singleWordLen);
                differ.put(word, differ.getOrDefault(word, 0) + 1);
            }

            for (String word : words) {
                differ.put(word, differ.getOrDefault(word, 0) - 1);
                if (differ.get(word) == 0) {
                    differ.remove(word);
                }
            }

            for (int start = i; start < sLen - singleWordLen * wordsLen + 1; start += singleWordLen) {

                if (start != i) {
                    String word = s.substring(start + (wordsLen - 1) * singleWordLen, start + wordsLen * singleWordLen);
                    differ.put(word, differ.getOrDefault(word, 0) + 1);
                    if (differ.get(word) == 0) {
                        differ.remove(word);
                    }
                    word = s.substring(start - singleWordLen, start);
                    differ.put(word, differ.getOrDefault(word, 0) - 1);
                    if (differ.get(word) == 0) {
                        differ.remove(word);
                    }
                }
                if (differ.isEmpty()) {
                    res.add(start);
                }
            }

        }

        return res;

    }
}