question:

Give the order of growth (as a function of N) of the running times of each of the following code fragments:

a.

int sum = 0;
for(int n = N; n > 0; n /= 2)
    for(int i = 0; i < n; i++)
       sum++;

b.

int sum = 0;
for(int i = 1; i < N; i*=2)
    for(int j = 0; j < i; j++)
       sum
 
++;

c.

int sum = 0;
for(int i = 1; i < N; i*=2)
    for(int j = 0; j < N; j++)
       sum++;

answer:

a. N + N/2 + N/4 + ... 即N（等比求和）

b. 1 + 2 + 4 + 8 + ... 即N //和a一模一樣，只是順序反了

c. 外層lgN， 內層N，且互不影響，即NlgN

//官網答案

Answer: linear (N + N/2 + N/4 + ...); linear (1 + 2 + 4 + 8 + ...); linearithmic (the outer loop loops lg N times).

1.4.6