[LeetCode]Substring with Concatenation of All Words

Substring with Concatenation of All Words

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Question

Total Accepted: 42899 Total
Submissions: 215720 Difficulty: Hard

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 wordsexactly once and without any intervening characters.
For example, given:

s:

"barfoothefoobarman"

words:

["foo", "bar"]

You should return the indices:

[0,9]

.

(order does not matter).

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这个题目是一个非常让人蛋疼的事情，提交了好多次都是超时。尽量在各个地方节约时间

上代码

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;

public class Solution {

public List<Integer> findSubstring(String s, String[] words) {
List<Integer>result = new ArrayList<>();
HashMap<String, Integer>hashMap=new HashMap<>();
HashMap<String, Integer>curHashMap=new HashMap<>();
if(words.length==0||s.length()<words[0].length())return result;
int len=words[0].length();
for(int i=0;i<words.length;i++){
Integer count = hashMap.get(words[i]);
if(count==null){
hashMap.put(words[i], 1);
}
else{
hashMap.put(words[i],count+1);
}
}
//因为题目中要求words中所有的词都出现一变，所以i的范围要控制，减小循环次数
for(int i = 0;i<=s.length()-len*words.length;i++){
int j;
curHashMap.clear();
for(j=0;j<words.length;j++){
String word = s.substring(i+j*len,i+(j+1)*len);
if(!hashMap.containsKey(word)){
break;
}
Integer count = curHashMap.get(word);
if (count==null) {
curHashMap.put(word, 1);
}
else{
curHashMap.put(word,count+1);//这里用的是+号，如果用hashmap-1的话会超时
}
if(curHashMap.get(word)>hashMap.get(word)){
break;
}
}
if(j==words.length){
result.add(i);
}

}

return result;
}

public static void main(String[]args){
Solution solution = new Solution();
System.out.println(solution.findSubstring("aaaa",new String[]{"aa","aa"}));
}
}