思路：搜索分割点，每次判断当前分割点喝上一个分割点之间的数据是否为回文串，如果是，就继续搜索下一个。
（来实验室工作的第二天，加油！）

class Solution {
    String str;
    int len;
    LinkedList<LinkedList<Integer>> list = new LinkedList<>();

    public boolean check(String s){
        // 判断s是否是回文字符串
        if(s.length()==0 || s.length()==1){
            return true;
        }
        return s.charAt(0)==s.charAt(s.length()-1) && check(s.substring(1,s.length()-1));
    }
    
    public void dfs(int splitIndex, LinkedList<Integer> tmp){
        if(splitIndex==len+1){
            list.add(new LinkedList<>(tmp));
            return;
        }

        for(int i=splitIndex; i<len+1;i++){
            int lastSplitIndex = 0;
            if(tmp.size() != 0){
                lastSplitIndex = tmp.get(tmp.size()-1);
            }
            String subStr = str.substring(lastSplitIndex, i);
            if(check(subStr)){
                tmp.add(i);
                dfs(i+1, tmp);
                tmp.removeLast();
            }
        }

        return;
    }

    public List<List<String>> partition(String s) {
        // 遍历切割点即可
        this.str = s;
        this.len = str.length();

        LinkedList<Integer> tmp = new LinkedList<>();
        dfs(1, tmp);    //遍历切割点

        // 将分割点转为字符串
        List<List<String>> res = new LinkedList<>();
        for(LinkedList<Integer> item: list){
            List<String> res_item = new LinkedList<>();
            int lastSplitIndex = 0;
            for(int currentSplitIndex : item){
                String subStr = str.substring(lastSplitIndex, currentSplitIndex);
                res_item.add(subStr);
                lastSplitIndex = currentSplitIndex;
            }
            res.add(res_item);
        }

        return res;
    }
}