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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 should return the indices:
[0,9].
(order does not matter).
算法思路：
1. 遍历S，每遇到一个字符，就取出wordLength的长度，并验证是否在L中，如不在，i++，如在，进行loop迭代，直到找到一个concatenation （记录下来）或者失配。继续i++，时间复杂度最好为O(n)，最坏为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 }
思路2：
优化，双指针法，思想类似 Longest Substring Without Repeating Characters， 复杂度可以达到线性。（待验证）
参考链接：
http://blog.csdn.net/ojshilu/article/details/22212703
http://blog.csdn.net/linhuanmars/article/details/20342851
[leetcode]Substring with Concatenation of All Words,布布扣,bubuko.com
[leetcode]Substring with Concatenation of All Words
标签：style blog http color os strong io art
原文地址：http://www.cnblogs.com/huntfor/p/3866125.html
