leetcode 30 Substring with Concatenation of All Words

lc30 Substring with Concatenation of All Words

两个hashmap

一个用来记录words[]中每种单词的出现次数，用来之后做匹配

一个用来记录source字符串i~j中每种单词出现次数，与前者比对

思路就是检查source字符串所有连续的长为words[].length()*words[0].length的子串

for(i=0; i<s.length() - words.length()*words[0].length; i++)

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        if(s.length() == 0 || words.length == 0)
            return new ArrayList<Integer>();
        HashMap<String, Integer> count = new HashMap<>();
        
        for(String i : words)
            count.put(i, count.getOrDefault(i, 0) + 1);
        int len = words[0].length();
        int wordsNum = words.length;
        List<Integer> res = new ArrayList<>();
        
        for(int i=0; i<s.length() - len*wordsNum + 1; i++){
            HashMap<String, Integer> seenWords =  new HashMap<>();
            int j = 0;
            while(j < wordsNum){
                String word = s.substring(i + j*len, i + (j+1)*len);
                if(count.containsKey(word)){
                    seenWords.put(word, seenWords.getOrDefault(word, 0) + 1);
                    if(seenWords.get(word) > count.get(word))
                         break;
                
                }else
                    break;
                j++;
            }
            if(j == wordsNum){
                res.add(i);
            }
        }
        return res;
    }
}