# [Leetcode] substring with concatenation of all words

Source: Internet
Author: User

Substring with concatenation of all words

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

For example, given:
S:`"barfoothefoobarman"`
L:`["foo", "bar"]`

You shoshould return the indices:`[0,9]`.
(Order does not matter ).

Algorithm ideas:

1. traverse S, get the length of wordlength every time a character is encountered, and verify whether it is in L, if not, I ++, such as, loop iteration, until a concatenation (recorded) or mismatch is found. Continue with I ++. The best time complexity is O (n), and the worst is O (n * num * wordlength)

` 1 public class Solution { 2     List<Integer> list = new ArrayList<Integer>(); 3     public List<Integer> findSubstring(String S, String[] L) { 4         int num = L.length; 5         int wordLength = L[0].length(); 6         if(S.length() < wordLength * num) return list; 7         HashMap<String,Integer> hash = new HashMap<String,Integer>(); 8         for(int i = 0; i < L.length; i++){ 9             if(hash.containsKey(L[i])) hash.put(L[i],hash.get(L[i]) + 1);10             else hash.put(L[i],1);11         }12         HashMap<String,Integer> copy = new HashMap<String,Integer>(hash);13         for(int i = 0; i <= S.length() - wordLength; i++){14             int start = i;15             int end = start + wordLength;16             String sub = S.substring(start, end); 17             if(copy.get(sub)!=null && copy.get(sub) != 0){18                 int count = 0;19                 boolean canLoop = true;20                 while(canLoop){21                     copy.put(S.substring(start, end), copy.get(S.substring(start, end)) - 1);22                     count++;23                     if(count == num){24                         list.add(i);25                         copy = (HashMap<String,Integer>)hash.clone();26                         break;27                     }28                     start = end;29                     end += wordLength;30                     if(end > S.length() || !copy.containsKey(S.substring(start, end)) || copy.get(S.subSequence(start, end).toString()) <= 0){31                         canLoop = false;32                         copy = (HashMap<String,Integer>)hash.clone();33                     }34                 }35             }36         }37         return list;38     }39 }`

Idea 2:

Optimization, double pointer method, similar to longest substring without repeating characters, the complexity can reach linear. (To be verified)

Http://blog.csdn.net/ojshilu/article/details/22212703

Http://blog.csdn.net/linhuanmars/article/details/20342851

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.