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).

 

从左往右扫描主字符串，用两个map<String, Integer>来统计字符串的出现次数. 这里L数组中可能会出现两个相同的字符串. 

 

public class Solution {
    public ArrayList<Integer> findSubstring(String S, String[] L) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        if(S == null || S.length() == 0 || L.length == 0){
            return null;
        }
        
        int numWords = L.length;
        int M = S.length();
        int N = L[0].length();
        HashMap<String, Integer> expected = new HashMap<String, Integer>();
        HashMap<String, Integer> real = new HashMap<String, Integer>();
        
        for(int i = 0; i < L.length; i++){
            if(expected.get(L[i]) == null){
                expected.put(L[i],1);
                real.put(L[i], 0);
            }else{
                expected.put(L[i], expected.get(L[i])+1);
            }
        }
        // i的遍历范围是S的长度，减去L中所有String的长度和
        for(int i = 0; i <= (M-N*numWords); i++){
            int j  = 0;
            while(j<numWords){
                // 这里的j*N，j表示已经遇到的String的个数，N是L中每个String的长度 
                String tmp = S.substring(i+j*N, i+(j+1)*N);
                if(expected.get(tmp) == null){
                    break;
                }else{
                    real.put(tmp, real.get(tmp) +1);
                    // 下面的if 表示如果遇到String超过 expected了 就跳出
                    if(real.get(tmp) > expected.get(tmp)){
                        break;
                    }
                    j++;
                }
            }
            // j 表示在S中遇到L中String的个数，j == len 表示全部都遇到了
            if(j == numWords){
                res.add(i);
            }
            // 这里是将real清零 重新计算遇到的在L中的String
            for(int k= 0; k < L.length; k++){
                real.put(L[k], 0);
            }
        }
        return res;
    }
}