List中取出相同的值 组成一个个小的list

首先说明，使用如下方法是不可以的：

 List list=new ArrayList<>();
      for(int i = 0; i < ll.size(); i++){
          list.add(ll.get(i));
          ll.remove(i);
        }

说明使用remove是不可行的，因为remove会删除i位置的元素，然后后面的元素统一向前移一位，这样你会发现，得到元素莫名的少了

废话不多说，首先看我的需求，

我有一个实体类，结构如下：

 {
      "start_date": 691200, 
      "start_time": "00:00", 
      "content": "哈哈"
    }, 

完整的json太多，列出一部分：

"routes": [
    {
      "start_date": 0, 
      "start_time": "00:00", 
      "content": "5oiQ6YO96ZuG5ZCI"
    }, 
    {
      "start_date": 0, 
      "start_time": "00:00", 
      "content": "6aSQ6aOfOuWFqOWkqemkkOi0ueiHqueQhg=="
    }, 
    {
      "start_date": 0, 
      "start_time": "00:00", 
      "content": "5L2P5a6/OuaIkOmDveW4guWMuu+8jOmFkuW6lzItM+S6uuagh+WHhumXtA=="
    }, 
    {
      "start_date": 86400, 
      "start_time": "00:00", 
      "content": "5oiQ6YO9LembheWuiS3kuozpg47lsbEt5rO45a6aLeW6t+Wumi3mlrDpg73moaUgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgICAgIOOAgg=="
    }, 
    {
      "start_date": 86400, 
      "start_time": "00:00", 
      "content": "5L2P5a6/OuaWsOmDveahpeW9k+WcsOWuvummhu+8iOagh+mXtO+8iQ=="
    }, 
    {
      "start_date": 86400, 
      "start_time": "00:00", 
      "content": "6aSQ6aWuOuazuOWumuS4remkkCzmlrDpg73moaXmmZrppJA="
    }, 
    {
      "start_date": 86400, 
      "start_time": "00:00", 
      "content": "6ZyA5pe2OuaIkOmDvS3lurflrprnuqY2LjXlsI/ml7bvvIzlurflrpot5paw6YO95qGl57qmMy415bCP5pe2"
    }, 
    {
      "start_date": 86400, 
      "start_time": "00:00", 
      "content": "5rW35ouUOuaIkOmDvTU0MOexs++8jOW6t+WumjI5OTDnsbPvvIzmipjlpJrlsbE0Mjk457Gz77yM5paw6YO95qGlMzQwMOexsw=="
    },     
    {
      "start_date": 259200, 
      "start_time": "00:00", 
      "content": "5a2Q5qKF5p2RLei0oeWYjuWvui3otKHlmI7lhrDlt50t5aSn5pys6JCl"
    }, 
    {
      "start_date": 259200, 
      "start_time": "00:00", 
      "content": "6aSQ6aWuOuWtkOaiheadkeaXqemkkCDoh6rlpIfot6/ppJAg6JCl5Zyw5pma6aSQ"
    }, 
    {
      "start_date": 259200, 
      "start_time": "00:00", 
      "content": "5L2P5a6/OuWkp+acrOiQpQ=="
    }, 
    {
      "start_date": 259200, 
      "start_time": "00:00", 
      "content": "5rW35ouUOui0oeWYjuWvujM4NTBt44CB6JCl5ZywNDIwMG0="
    }, 
    {
      "start_date": 345600, 
      "start_time": "00:00", 
      "content": "5aSn5pys6JClLemAguW6lOaAp+iuree7gy1DMe+8iDQ5MDDlsbHlnbPokKXlnLDvvIk="
    }, 
    {
      "start_date": 345600, 
      "start_time": "00:00", 
      "content": "6aSQ6aWuOuiQpeWcsOaXqemkkOOAgei3r+mkkOOAgeiQpeWcsOaZmumkkA=="
    }, 
    {
      "start_date": 345600, 
      "start_time": "00:00", 
      "content": "5L2P5a6/OumcsuiQpQ=="
    }, 
    {
      "start_date": 345600, 
      "start_time": "00:00", 
      "content": "6ZyA5pe2OkJDLUMx57qm6ZyANOWwj+aXtg=="
    }, 
    {
      "start_date": 345600, 
      "start_time": "00:00", 
      "content": "5rW35ouUOkMx5rW35ouU57qmNDkwMOexsw=="
    }   
  ]

就像这样的，我需要把start_date相同的组成一个新的list，但是有多少个list，每个list的长度是多少，都不确定，这样我们的思路还是取出一个删除一个，但是不能使用remove，这样会有问题，那么怎么删除呢？这就要用到Iterator（迭代），官方给的迭代器就可以完成这种操作，它的原理大致是：讲list拷贝一份给Iterator迭代器，然会便利迭代器，便利的时候边取值，边删除，删除完成后在更新list，这样list就不会因为同时取值和删除二有问题啦，好啦，看代码：

private void setDataToLv(List<RoutesBean> list){
        List<List<RoutesBean>> ll=new ArrayList<>();
        while (list.size()>0){
            RoutesBean b=list.get(0);
            List<RoutesBean>  l=new ArrayList<>();
            Iterator iterator=list.iterator();
            while (iterator.hasNext()){
                RoutesBean bean= (RoutesBean) iterator.next();
                if (bean.getStartDate()==b.getStartDate()){
                    l.add(bean);
                    iterator.remove();
                }
            }
            if (l.size()>0)
            ll.add(l);
        }

        //打印测试
        for (int i = 0; i < ll.size(); i++) {
            Log.e("qqqqq",ll.get(i).get(0).getStartDate()+"=="+ll.get(i).size());
        }

    }