[英]What's the Big O time complexity of this code?
#include <stdio.h>
int F(int L[], int p, int q) {
if (p < q) {
int r, f1, f2;
r = (p + q) / 2;
f1 = 2 * F(L, p, r);
f2 = 2 * F(L, r + 1, q);
return f1 + f2;
} else if (p == q) {
return L[p] * L[p];
}else{
return 0;
}
}
int main(void) {
int arr[8] = {1,2,3,4,5,6,7};
printf("%d", F(arr, 0, 7));
}
Someone said the time complexity of this code is O(n)
.有人说这段代码的时间复杂度是O(n)
。
I don't understand it at all...我完全不明白...
Isn't it O(logN)
???不是O(logN)
吗???
Answer: The Big-O complexity is O(N)答案:Big-O 复杂度为 O(N)
Explanation:解释:
The program takes a range of some size (ie q - p + 1
) and cut that range into 2 half.该程序采用一定大小的范围(即q - p + 1
)并将该范围切成两半。 Then it calls the function recursively on these two half ranges.然后它在这两个半范围内递归调用 function。
That process continues until the range has size 1 (ie p == q
).该过程一直持续到范围的大小为 1(即p == q
)。 Then there is no more recursion.然后就没有递归了。
Example: Consider a start range with size 8 (eg p=0, q=7
) then you will get示例:考虑大小为 8 的起始范围(例如p=0, q=7
),然后您将得到
1 call with range size 8
2 calls with range size 4
4 calls with range size 2
8 calls with range size 1
So 7 calls (ie 1+2+4) with range size greater than 1 and 8 calls with range size equal to 1. A total of 15 calls which is nearly 2 times the starting range size.因此,范围大小大于 1 的 7 个调用(即 1+2+4)和范围大小等于 1 的 8 个调用。总共 15 个调用,几乎是起始范围大小的 2 倍。
So for a range size being a power of 2, you can generalize to be因此,对于范围大小为 2 的幂,您可以概括为
Number of calls with range size greater than 1:
1+2+4+8+16+...+ rangesize/2 = rangesize - 1
Number of calls with range size equal to 1:
rangesize
So there will be exactly 2 * rangesize - 1
function calls when range size is a power of 2.因此,当范围大小为 2 的幂时,将恰好有2 * rangesize - 1
function 次调用。
That is Big-O complexity O(N).那就是 Big-O 复杂度 O(N)。
Want to try it out?想尝试一下吗?
#include <stdio.h>
unsigned total_calls = 0;
unsigned calls_with_range_size_greater_than_one = 0;
unsigned calls_with_range_size_equal_one = 0;
int F(int L[], int p, int q) {
++total_calls;
if (p < q) {
++calls_with_range_size_greater_than_one;
int r, f1, f2;
r = (p + q) / 2;
f1 = 2 * F(L, p, r);
f2 = 2 * F(L, r + 1, q);
return f1 + f2;
} else if (p == q) {
++calls_with_range_size_equal_one;
return L[p] * L[p];
}else{
return 0;
}
}
int arr[200] = {1,2,3,4,5,6,7};
int main(void) {
for (int i=3; i < 128; i = i + i + 1)
{
total_calls=0;
calls_with_range_size_greater_than_one=0;
calls_with_range_size_equal_one=0;
F(arr, 0, i);
printf("Start range size: %3d -> total_calls: %3u calls_with_range_size_greater_than_one: %3u calls_with_range_size_equal_one: %3u\n", i+1, total_calls, calls_with_range_size_greater_than_one, calls_with_range_size_equal_one);
}
return 0;
}
Output: Output:
Start range size: 4 -> total_calls: 7 calls_with_range_size_greater_than_one: 3 calls_with_range_size_equal_one: 4
Start range size: 8 -> total_calls: 15 calls_with_range_size_greater_than_one: 7 calls_with_range_size_equal_one: 8
Start range size: 16 -> total_calls: 31 calls_with_range_size_greater_than_one: 15 calls_with_range_size_equal_one: 16
Start range size: 32 -> total_calls: 63 calls_with_range_size_greater_than_one: 31 calls_with_range_size_equal_one: 32
Start range size: 64 -> total_calls: 127 calls_with_range_size_greater_than_one: 63 calls_with_range_size_equal_one: 64
Start range size: 128 -> total_calls: 255 calls_with_range_size_greater_than_one: 127 calls_with_range_size_equal_one: 128
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