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 should return the indices: [0,9]
.
(order does not matter).
public class Solution { public List<Integer> findSubstring(String S, String[] L) { List<Integer> list = new ArrayList<Integer>(); if(L.length == 0){ return list; } int lslen = L[0].length(); int inlen = lslen*L.length; if(S.length()<inlen){ return list; } Map<String,Integer> map = new HashMap<>(); for(int i =0;i<L.length;i++){ if (map.containsKey(L[i])){ map.put(L[i], map.get(L[i])+1); }else{ map.put(L[i], 1); } } for(int i =0;i<=S.length()-inlen;i++){ Map<String,Integer> hmap = new HashMap<>(map); for(int j = i;j<=S.length()-lslen;j+=lslen){ String temp = S.substring(j,j+lslen); if(!hmap.containsKey(temp)) break; if(hmap.get(temp)==1){ hmap.remove(temp); }else{ hmap.put(temp, hmap.get(temp)-1); } } if(hmap.size()==0){ list.add(i); } } return list; } }
[LeetCode]Substring with Concatenation of All Words
原文地址:http://blog.csdn.net/guorudi/article/details/39234945