-
But if, at any point, I get to a place in the list where the thing I'm looking for is smaller than the element in the list, I know everything else in the rest of the list has to be bigger than that, I don't have to bother looking anymore.
比当前位置数组的元素要小,我也就知道后面的数肯定,也都比我的目标数要大了，我就不用再继续进行下去了,这意味着目标数不在这个数组中，我就可以退出了。
-
If it's there, I'm done, if not, I keep walking down, and I only stop when I get to a place where the element I'm looking for is smaller than the value in the list., in which case I know the rest of this is too big and I can stop.
并且保持遍历,我只在当当前位置的数组元素,大于目标数时停止,这意味着剩下的元素都比目标元素大，但是其他的情况,我还是要遍历完整个数组。
-
The easy thing to do would be the following: start at the front end of the list, check the first element. If it's the thing I'm looking for, I'm done. It's there. If not, move on to the next element. And keep doing that.
从数组的第一个元素开始：,如果这是目标元素那么结束,如果不是的话就继续比较下面的元素，并一直这么比较下去,但如果，在某一点我发现我要寻找的数字。
-
And that's like kind of hoping that the element you're searching for is the first in the list and the last in the list. Maybe.
也就是希望要找的元素,在数组的头部或者尾部,情况可能是这样的。
-
That one's not so obvious. So let's think about this for a second. To sort a list in linear time, would say, I have to look at each element in the list at most a constant number of times.
所以让我们来思考一会,要在线性时间能排序，列表里每个元素最多被使用常数次,不一定是一次，对吧。
-
I'm accessing a list. How long does it take for me to get the nth element of a list?
我取得数组的第N个元素,需要多长时间呢？
-
I'll let you just grok it but you can see it's basically doing what I did over there. Setting up two indices for the two sub-list,it's just walking down, finding the smallest element, putting it into a new list. When it gets to the end of one of the lists, it skips to the next part, and only one of these two pieces will get called because only one of them is going to have things leftovers.
你们可以大体的浏览一下,但是它们基本就是我在那里所做的事情,为两个子列表设置了两个指针,指针顺着列表走下去，找到最小的元素,把它放入到一个新的列表中去,当它走到一个列表的尾部时,它会跳到下部分去,两部分中只有一个会被执行,因为只有一个会有元素剩余。
应用推荐