[LeetCode] Substring with Concatenation of All Words

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).
这两天为了准备WAP的面试正好在练习JAVA,于是重写了这道题，发现JAVA果然好简洁。

思路就不说了，还是那个滑动窗口。

public class Solution {
public ArrayList<Integer> findSubstring(String S, String[] L) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList<Integer> ret=new ArrayList<Integer>();
if ( S==null || L==null || L.length==0 )
return ret;
HashMap<String,Integer> words=new HashMap<String,Integer>();
// HashMap<String,Integer> need =new HashMap<String,Integer>();
for(int i=0;i<L.length;i++){
if ( words.containsKey(L[i]) ){
int t = words.get(L[i]);
words.put(L[i], t+1);
//need.put(L[i], t+1);
}else{
words.put(L[i],1);
//	need.put(L[i],1);
}

}

int wordLength=L[0].length();
for(int i=0;i<wordLength;i++){
HashMap<String,Integer> need =new HashMap<String,Integer>(words);
int start=i,end=i;
int needCnt=L.length;
if ( (S.length()-start)< (wordLength*needCnt))
break;

for(;end<S.length();end+=wordLength){
if ( S.length()-end < wordLength )
break;
String tmp= S.substring(end,end+wordLength);
if (words.containsKey(tmp)){
int t=need.get(tmp);
need.put(tmp,t-1);
if ( t>0)
needCnt--;
}

if ( needCnt==0 ){
while(true){
String toRemove=S.substring(start,start+wordLength);
if (words.containsKey(toRemove) ){
int t =need.get(toRemove);
if ( t <0 ){
need.put(toRemove,t+1);
}
else
break;
}
start+=wordLength;
}
if (end-start == wordLength*(L.length-1))
ret.add(start);
}
}
}
return ret;
}
}