【leetcode每日刷题】30. Substring with Concatenation of All Words

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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 "barfoor" 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: []

import java.util.ArrayList;
import java.util.Arrays;

class num30 {
    public ArrayList<Integer> findSubstring(String s, String[] words) {
        int lenarr = words.length;
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(lenarr == 0) return result;
        int lenStr = words[0].length();
        for(int i=0; i <= s.length() - lenStr*lenarr; i++){
            ArrayList<String> temp = new ArrayList<String>(Arrays.asList(words));
            for(int j=0; j<lenarr; j++){
                int pos = temp.indexOf(s.substring(i+j*lenStr, i+(j+1)*lenStr));
                if(pos == -1) break;
                temp.remove(pos);
            }
            if(temp.size() == 0) result.add(i);
        }
        return result;
    }
    public static void main(String[] args) {
        num30 n = new num30();
        String s = "barfoothefoobarman";
        String[] words = new String[]{"foo", "bar"};
        ArrayList<Integer> result = n.findSubstring(s, words);
        for(int i=0; i<result.size(); i++){
            System.out.println(result.get(i));
        }
    }
}

因为每个单词都是相同的长度，所以只需要遍历字符串长度中的0到 s.length() - lenStr*lenarr部分，本例中是s.length-2*3

然后截取lenarr次 [ i, i+j*lenstr ]的长度，如果在words中，则删除words中的这个词，最后如果都找到，则添加起始索引。