30. Substring with Concatenation of All Words

最新推荐文章于 2024-01-06 14:27:05 发布

原创最新推荐文章于 2024-01-06 14:27:05 发布 · 149 阅读

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本文介绍了一种寻找字符串中特定子串的方法，特别是在一个长字符串中找出由多个固定长度单词组成的子串的所有起始索引。通过使用哈希表和Boyer-Moore算法，该方法能够有效地解决这一问题。

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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 words exactly 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).

做法挺暴力的，就是一个个位置扫，检查从每个位置开始是否包含了words集合里面的所有串刚好一边。

整体可以用hash加速，字串匹配可以上Boyer-Moore

public class Solution {
   public List<Integer> findSubstring(String s, String[] words)
	{
		List<Integer> arraylist=new ArrayList<>();
		
		if(s.length()==0)
			{
				for(String word : words)
					if(word.length()>0)
						return arraylist;
				arraylist.add(0);
				return arraylist;
			}
		
		int wordlen=words[0].length();
		int size=words.length;
		
		
		int R = 128;
		int[][] right=new int[size][R];
		
		for(int i=0;i<size;i++)
			for (int c = 0; c < R; c++)
				right[i][c] = -1;
		
		for(int i=0;i<size;i++)
			for (int j = 0; j < words[i].length(); j++)
				right[i][words[i].charAt(j)] = j;
		
		HashSet<Integer> hashset=new HashSet<>(20000);
		HashSet<Integer> matchhashset=new HashSet<>(6000);

		
L1:		for(int i=0;i+size*wordlen<=s.length();i++)
		{
			String checknow=s.substring(i, i+size*wordlen);
			int hash=checknow.hashCode();
			if(!hashset.contains(hash))
				hashset.add(hash);
			else {
				if(matchhashset.contains(hash))
					arraylist.add(i);
				continue;
			}
			
			boolean[] find=new boolean[size];
			for(int j=1;j<=size;j++)
			{
				int start=i+(j-1)*wordlen;
				int end=i+j*wordlen;
				String sub=s.substring(start, end);
				boolean success=false;
				for(int m=0;m<size;m++)
					if(!find[m])
					{
						if(search(sub, words[m], right[m])==true)
						{
							find[m]=true;
							success=true;
							break;
						}
					}
				if(!success)
					continue L1;
			}
			arraylist.add(i);
			matchhashset.add(hash);
		}
		
		for(String word:words)
		{
			if(word.length()>0)
				return arraylist;
		}
		arraylist.add(s.length());
		
		return arraylist;
		
	}
	
	public boolean search(String txt,String pat,int[] right)
	{
		
		int M = pat.length();
		int N = txt.length();
		
		int skip;
		for (int i = 0; i <= N - M; i += skip)
		{
			skip = 0;
			for (int j = M - 1; j >= 0; j--)
				if (pat.charAt(j) != txt.charAt(i + j))
				{
					skip = j - right[txt.charAt(i + j)];
					if (skip < 1)
						skip = 1;
					break;
				}
			if (skip == 0)
				return true;
		}

		return false;
	}
}