現假設有下面這個List：
List<Integer> list = new ArrayList<>();list.add(1);list.add(2);list.add(3);list.add(4);如果我想把 2 和 3 這兩個元素從該List中移除，大家的第一想法可能是：
list.remove(1);// 此處為啥還是 1 ，這個應該不用解釋了吧list.remove(1);這樣做，當數據量比較少時，是沒有問題的，但是當數據量多的時候，就很麻煩了。 有人可能會想說，數據量多的時候，可以使用 for 循環來做啊：
for(int i=0; i<2; i++) { list.remove(1);}OK，這樣也是可以的。 不過，下面我想介紹一種更直觀的方式。直接通過List的API來實現：
list.subList(1,3).clear();這個API是不是一目了然呢：
subList(1, 3)： 取從 第1(包含) 到 第3(不包含) 的這幾個元素的子集clear()： 將這個子集清空為什么這么做是可以的呢？請看下面 subList() 這個API的 官方說明：
Returns a view of the portion of this list between the specified fromIndex, inclusive, and toIndex, exclusive. (If fromIndex and toIndex are equal, the returned list is empty.) The returned list is backed by this list, so non-structural changes in the returned list are reflected in this list, and vice-versa. The returned list supports all of the optional list Operations supported by this list.
This method eliminates the need for explicit range operations (of the sort that commonly exist for arrays). Any operation that expects a list can be used as a range operation by passing a subList view instead of a whole list. For example, the following idiom removes a range of elements from a list:
list.subList(from, to).clear();Similar idioms may be constructed for indexOf and lastIndexOf, and all of the algorithms in the Collections class can be applied to a subList.
The semantics of the list returned by this method become undefined if the backing list (i.e., this list) is structurally modified in any way other than via the returned list. (Structural modifications are those that change the size of this list, or otherwise perturb it in such a fashion that iterations in PRogress may yield incorrect results.)
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