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What Faulkner saw was that there could be no order at all, no idea of doing what is right, in a world that measured success in terms of money.
VOA: special.2010.01.03
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OK. What would you guess the order of growth here is? Yeah. Why? Good. Exactly. Right?
如果没听到答案的同学,答案是对数级的？
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All right, so the bonding order, you're correct, should be 2, if we subtract the number of bonding minus anti-bonding electrons and take that in 1/2.
好，你们是对的，键序为，如果我们用成键数,减去反键数除以2。
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We said each of the merge operations O was of order n. But n is different. Right?
注意这里发生了什么,我们说过每一次合并操作的复杂度都是？
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Just as in grade school when you're doing division and multiplication, you do it from left to right in terms of order of operations.
正像是在小学你做除法和乘法的时候,你毫不含糊地从左到右,按顺序做运算。
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You may know from reading food labels that these ingredients in any food label are listed in order of how much there is in the food itself, so sugar comes right after peanuts.
看看食品标签你就会知道,食物配料都列在食品标签上,以它在食品中的含量多少的顺序排列,所以糖的含量仅次于花生
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Right. Order of growth here.
你们是怎么想的？
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This is a game where you get a little plastic device, it's got 15 plastic numbers on it and one hole in this little plastic board and you can move those numbers up, down, left to right, and the goal is to take what's a random assortment of tiles and arrange them in numeric order and that's one of these little things you can play sort of absentmindedly.
在这个游戏中，你会拿到一个可塑的装置，上面有15个数字和一个孔,你可以将数字上下移动，左右移动，目标就是,拿到一个随机的组合后,将其按数字顺序排列,这个你可以随便玩玩。
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In the linear case, meaning in the unsorted case what's the complexity of this? k times n, right? Order n to do the search, and I've got to do it k times, so this would be k times n.
复杂度是多少？k的n次方，对吧？,在序列n中做搜索，要做k次，所以是k的n次方次,如果先排序后搜索。
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