LeetCode 30. Substring with Concatenation of All Words

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这题做了好几次都超时。答案是看的别人的博客学习的。。
public class Solution {    public List<Integer> findSubstring(String s, String[] words) {       List<Integer> res = new LinkedList<Integer>();        if(words == null || words.length == 0 || s == null || s.equals("")) return res;        HashMap<String, Integer> freq = new HashMap<String, Integer>();        // 统计数组中每个词出现的次数，放入哈希表中待用        for(String word : words){            freq.put(word, freq.containsKey(word) ? freq.get(word) + 1 : 1);        }        // 得到每个词的长度        int len = words[0].length();        // 错开位来统计        for(int i = 0; i < len; i++){            // 建一个新的哈希表，记录本轮搜索中窗口内单词出现次数            HashMap<String, Integer> currFreq = new HashMap<String, Integer>();            // start是窗口的开始，count表明窗口内有多少词            int start = i, count = 0;            for(int j = i; j <= s.length() - len; j += len){                String sub = s.substring(j, j + len);                // 看下一个词是否是给定数组中的                if(freq.containsKey(sub)){                    // 窗口中单词出现次数加1                    currFreq.put(sub, currFreq.containsKey(sub) ? currFreq.get(sub) + 1 : 1);                    count++;                    // 如果该单词出现次数已经超过给定数组中的次数了，说明多来了一个该单词，所以要把窗口中该单词上次出现的位置及之前所有单词给去掉                    while(currFreq.get(sub) > freq.get(sub)){                        String leftMost = s.substring(start, start + len);                        currFreq.put(leftMost, currFreq.get(leftMost) - 1);                        start = start + len;                        count--;                    }                    // 如果窗口内单词数和总单词数一样，则找到结果                    if(count == words.length){                        String leftMost = s.substring(start, start + len);                        currFreq.put(leftMost, currFreq.get(leftMost) - 1);                        res.add(start);                        start = start + len;                        count--;                    }                // 如果截出来的单词都不在数组中，前功尽弃，重新开始                } else {                    currFreq.clear();                    start = j + len;                    count = 0;                }            }        }        return res;    }}
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