PrefixAverages1(X)
Input: X, a 1-D numerical array of size n
1) Let A = an empty 1-D numerical array of size n
2) For i = 0 to n-1
3) Let s = X[0]
4) For j = 1 to i
5) Let s = s + X[j]
6) End For
7) Let A[i] = s /(i+1)
8) End For
Output: An n-element array A of numbers such that A[i]
is the average of elements X[0],X[1], … ,X[i]
PrefixAverages2(X)
Input: X, a 1-D numerical array of size n
1) Let A = an empty 1-D numerical array of size n
2) Let s = 0
3) For i = 0 to n-1
4) Let s = s + X[i]
5) Let A[i] = s / (i+1)
6) End For
Output: An n-element array A of numbers such that A[i]
is the average of elements X[0],X[1], … ,X[i]
所以这是我到目前为止所知道的:
第一个算法使用第二个嵌套的 for 循环。第二种效率更高。首先,S 等于数组的第一个元素。在第二个中,S 等于 0。
我还缺少什么?任何帮助将不胜感激,谢谢!