Ordering<Item> o = new Ordering<Item>() {
@Override
public int compare(Item left, Item right) {
return Ints.compare(left.price, right.price);
}
};
return o.max(list);
It's as efficient as it can be: it iterates through the items of the list, and returns the first of the Items having the maximum price: O(n).
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…