30. Substring with Concatenation of All Words

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).

 

本题很难，先说思路，首先要把words存放在hashmap里面，key存放String，value存放出现个数。因为words的string长度是一样的，假设长度为len，遍历s的时候只要在[0,len-1] 为开始点，然后每次加上len长度就可以了。同时还需要设置一个counter的长度，用来比较是否和words的长度相等。接下来再做一个hashmap，用来存储每次遍历的len长度的string，如果在words的hashmap里面出现了，则长度+1，否则清零。

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> res = new ArrayList<Integer>();
        Map<String,Integer> map = new HashMap<>();
        int len = words[0].length();
        if(s.length()==0||words.length==0) return res;
        for(String word:words){
            map.put(word,map.getOrDefault(word,0)+1);
        }
        for(int i=0;i<len;i++){
            int left = i;
            int counter = 0;
            Map<String,Integer> smap = new HashMap<>();
            for(int j=i;j<=s.length()-len;j+=len){
                String str = s.substring(j,j+len);
                if(map.containsKey(str)){
                    smap.put(str,smap.getOrDefault(str,0)+1);
                    if(smap.get(str)<=map.get(str)){
                        counter++;
                    }else{
                        while(smap.get(str)>map.get(str)){
                            String sstr = s.substring(left,left+len);
                            smap.put(sstr,smap.get(sstr)-1);
                            if(smap.get(sstr)<map.get(sstr)) counter--;
                            left+=len;
                        }
                    }
                    if(counter==words.length){
                        System.out.println(left);
                        res.add(left);
                        counter--;
                        String sstr = s.substring(left,left+len);
                        smap.put(sstr,smap.get(sstr)-1);
                        left = left+len;
                    }
                }else{
                    counter = 0;
                    left = j+len;
                    smap.clear();
                }
            }
        }
        return res;
    }
}