这是一个家庭作业的问题。他们说,需要 O(logN个+ logM),其中 N 和 M 是数组的长度。
This is a homework question. They say it takes O(logN + logM) where N and M are the arrays lengths.
让我们命名的数组 A 和 B 。显然,我们可以忽略所有 A [1] 和 B [I] 其中i> K。 首先让我们比较 A [K / 2] 和 B [K / 2] 。让 B [K / 2] > A [K / 2] 。因此,我们也可以放弃所有 B [I] ,其中i> K / 2。
Let's name the arrays a and b. Obviously we can ignore all a[i] and b[i] where i > k. First let's compare a[k/2] and b[k/2]. Let b[k/2] > a[k/2]. Therefore we can discard also all b[i], where i > k/2.
现在我们拥有所有 A [1] ,其中i< k和所有的 B [I] ,其中i<克/ 2中找到了答案。
Now we have all a[i], where i < k and all b[i], where i < k/2 to find the answer.
什么是下一个步骤?
推荐答案您已经知道了,只是坚持下去!而且要注意与索引...
You've got it, just keep going! And be careful with the indexes...
要简化一点,我会假设N和M是> K,所以在这里复杂度为O(日志K),这是为O(log N + M)的。
To simplify a bit I'll assume that N and M are > k, so the complexity here is O(log k), which is O(log N + log M).
伪code:
i = k/2 j = k - i step = k/4 while step > 0 if a[i-1] > b[j-1] i -= step j += step else i += step j -= step step /= 2 if a[i-1] > b[j-1] return a[i-1] else return b[j-1]有关演示,你可以使用循环不变I + J = K,但我不会做你的家庭作业:)
For the demonstration you can use the loop invariant i + j = k, but I won't do all your homework :)
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