所有的循環只為尋找答案, 所有的判斷只為選擇正確答案

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>();
        // 統計數組中每個詞出現的次數，放入哈希表中待用
        for(String word : words){
            freq.put(word, freq.containsKey(word) ? freq.get(word) + 1 : 1);
        }
        // 得到每個詞的長度
        int len = words[0].length();
        // 錯開位來統計
        for(int i = 0; i < len; i++){  // 因為要一單詞長度遞增, 因此i < len

            // 建一個新的哈希表，記錄本輪搜索中窗口內單詞出現次數
            HashMap<String, Integer> currFreq = new HashMap<String, Integer>();
            // start是窗口的開始，count表明窗口內有多少詞
            int start = i, count = 0;
            for(int j = i; j <= s.length() - len; j += len){
                String sub = s.substring(j, j + len);
                // 看下一個詞是否是給定數組中的
                if(freq.containsKey(sub)){
                    // 窗口中單詞出現次數加1
                    currFreq.put(sub, currFreq.containsKey(sub) ? currFreq.get(sub) + 1 : 1);
                    count++;
                    // 如果該單詞出現次數已經超過給定數組中的次數了，說明多來了一個該單詞，所以要把窗口中該單詞上次出現的位置及之前所有單詞給去掉
                    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--;
                    }
                    // 如果窗口內單詞數和總單詞數一樣，則找到結果
                    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--;
                    }
                // 如果截出來的單詞都不在數組中，前功盡棄，重新開始
                } else {
                    currFreq.clear();
                    start = j + len;
                    count = 0;
                }
            }
        }
        return res;
    }

30. Substring with Concatenation of All Words