给出一个数组。如何在恒定时间内找到索引间隔(i,j)中的元素总和。允许您使用多余的空间。 示例: 答:3 2 4 7 1 -2 8 0 -4 2 1 5 6 -1 长度= 14
Given an array. How can we find sum of elements in index interval (i, j) in constant time. You are allowed to use extra space. Example: A: 3 2 4 7 1 -2 8 0 -4 2 1 5 6 -1 length = 14
int getsum(int* arr, int i, int j, int len); // suppose int array "arr" is initialized here int sum = getsum(arr, 2, 5, 14);总和在恒定时间内应为10。
sum should be 10 in constant time.
推荐答案如果您可以花费O(n)时间来准备辅助信息,则可以基于该信息来计算O(1)中的总和,您可以轻松地做到这一点。
If you can spend O(n) time to "prepare" the auxiliary information, based on which you would be able calculate sums in O(1), you could easily do it.
准备(O(n)):
aux[0] = 0; foreach i in (1..LENGTH) { aux[i] = aux[i-1] + arr[i]; }查询(O(1)), arr 从 1 计算为 Length :
Query (O(1)), arr is numerated from 1 to LENGTH:
sum(i,j) = aux[j] - aux[i-1];我认为这是目的,因为否则,这是不可能的:对于任何 length 来计算 sum(0,length-1),您应该已经扫描了整个数组;至少需要线性时间。
I think it was the intent, because, otherwise, it's impossible: for any length to calculate sum(0,length-1) you should have scanned the whole array; this takes linear time, at least.
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如何在恒定时间内从给定的索引间隔(i,j)中找到元素的总和?
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