Leetcode Substring with Concatenation of All Words

最新推荐文章于 2019-12-26 08:25:23 发布

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本文介绍了一种使用HashMap实现的高效字符串匹配算法，该算法能够在给定字符串中找出由多个指定单词按顺序拼接而成的子串的所有起始位置。通过实例演示了如何利用队列和HashMap来跟踪单词出现的位置，并返回所有符合条件的索引。

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].

Difficulty: Hard

Solution:

HashMap<String, Queue<Integer>>

public class Solution {
    List<Integer> res = new ArrayList<Integer>();
    String[] wordList;
    HashMap<String, Queue<Integer>> map = new HashMap<String, Queue<Integer>>();
    
    public List<Integer> findSubstring(String s, String[] words) {
        if(words.length == 0) return res;
        int len = words[0].length();
        
        for(int i = 0; i < words.length; i++){
            if(map.containsKey(words[i])){
                map.get(words[i]).offer(-1);
            }
            else{
               Queue<Integer> q = new LinkedList<Integer>();
               q.offer(-1);
               map.put(words[i], q);
            }
        }
        wordList = new String[s.length() - len + 1];
        for(int i = 0; i < wordList.length; i++){
            wordList[i] = s.substring(i, i + len);
        }
        for(int i = 0; i < len; i++){
            int begin = 0, index = i, count = 0;
            while(begin < s.length()){
                // System.out.println(index);
                // System.out.println(index - len * count);
                // System.out.println("---");
                if(words.length == count){
                    //System.out.println("ADD");
                    res.add(index - len * count);
                    count--;
                    continue;
                }
                if((words.length - count) * len > s.length() - index){
                    break;
                }
                String newWord = wordList[index];
                if(!map.containsKey(newWord)){
                    index += len;
                    count = 0;
                    continue;
                }
                if(map.get(newWord).peek() >= index - count * len && map.get(newWord).peek() < index && (index - map.get(newWord).peek()) % len == 0){
                    int temp = map.get(newWord).poll();
                    map.get(newWord).offer(index);
                    count = (index - temp)/len;
                    index += len;
                    continue;
                }
                else{
                    map.get(newWord).poll();
                    map.get(newWord).offer(index);
                    index += len;
                    count++;
                }
            }
            if(words.length == count)
                res.add(index - len * count);
        }
        return res;
    }
}