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我目前正在大学学习数据结构,无意中遇到了一个关于递归复杂性的问题。
给定这段代码:
Unsigned func (unsigned n)
{
if (n ==0) return 1;
if(n==1) return 2;
\\do somthing(in O(1) )
return func(n-1) + func(n-1);
}
我知道代码的作用。我知道现在的形式,时间复杂度是O(2^n)。
但是我的问题是:如果我编写的代码不是最后一次返回调用,时间复杂度是否会改变:return 2*func(n-1)
?
我知道就内存复杂度而言,我们正在谈论递归占用空间的显着减少,但就时间复杂度而言,会有任何变化吗?
我使用递归函数进行了数学运算,发现时间复杂度不会发生变化,对吗?
最佳答案
这个方法只有 O(n)
,因为如果你用 5 运行它,它会用 4 递归,然后用 3 等等。
Unsigned func (unsigned n)
{
if (n ==0) return 1;
if(n==1) return 2;
\\do somthing(in O(1) )
return 2*func(n-1);
}
但是这个呢:
Unsigned func (unsigned n)
{
if (n ==0) return 1;
if(n==1) return 2;
\\do somthing(in O(1) )
return func(n-1) + func(n-1);
}
例如 func(5) 它将首先执行如下
5 -> 4 -> 3 -> 2 -> 1
然后它返回到 2,但是它会执行“第二”部分,所以整个过程看起来像
5 -> 4 -> 3 -> 2-> 1; 2-> 1; 3->2->1; etc.
因此它确实将复杂性从 O(n)
显着改变为 O(2^n)
让我们试试这个代码的实际例子:
public class Complexity {
private static int counter;
public static void main(String[] args) {
for (int i = 0; i < 20; i++) {
counter = 0;
func(i);
System.out.println("For n="+i+" of 2*func the number of cycles is " + counter);
counter = 0;
func2(i);
System.out.println("For n="+i+" of func + func the number of cycles is " + counter);
}
}
public static int func(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return 2 * func(n - 1);
}
public static int func2(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return func2(n - 1) + func2(n - 1);
}
}
有这样的输出:
For n=0 of 2*func the number of cycles is 1
For n=0 of func + func the number of cycles is 1
For n=1 of 2*func the number of cycles is 1
For n=1 of func + func the number of cycles is 1
For n=2 of 2*func the number of cycles is 2
For n=2 of func + func the number of cycles is 3
For n=3 of 2*func the number of cycles is 3
For n=3 of func + func the number of cycles is 7
For n=4 of 2*func the number of cycles is 4
For n=4 of func + func the number of cycles is 15
For n=5 of 2*func the number of cycles is 5
For n=5 of func + func the number of cycles is 31
For n=6 of 2*func the number of cycles is 6
For n=6 of func + func the number of cycles is 63
For n=7 of 2*func the number of cycles is 7
For n=7 of func + func the number of cycles is 127
For n=8 of 2*func the number of cycles is 8
For n=8 of func + func the number of cycles is 255
For n=9 of 2*func the number of cycles is 9
For n=9 of func + func the number of cycles is 511
For n=10 of 2*func the number of cycles is 10
For n=10 of func + func the number of cycles is 1023
For n=11 of 2*func the number of cycles is 11
For n=11 of func + func the number of cycles is 2047
For n=12 of 2*func the number of cycles is 12
For n=12 of func + func the number of cycles is 4095
For n=13 of 2*func the number of cycles is 13
For n=13 of func + func the number of cycles is 8191
For n=14 of 2*func the number of cycles is 14
For n=14 of func + func the number of cycles is 16383
For n=15 of 2*func the number of cycles is 15
For n=15 of func + func the number of cycles is 32767
For n=16 of 2*func the number of cycles is 16
For n=16 of func + func the number of cycles is 65535
For n=17 of 2*func the number of cycles is 17
For n=17 of func + func the number of cycles is 131071
For n=18 of 2*func the number of cycles is 18
For n=18 of func + func the number of cycles is 262143
For n=19 of 2*func the number of cycles is 19
For n=19 of func + func the number of cycles is 524287
但是如果您还记得已经计算出的结果,即使使用第二种方法,复杂度仍然是O(n)
:
public class Complexity {
private static int counter;
private static int[] results;
public static void main(String[] args) {
for (int i = 0; i < 20; i++) {
counter = 0;
func(i);
System.out.println("For n="+i+" of 2*func the number of cycles is " + counter);
counter = 0;
func2(i);
System.out.println("For n="+i+" of func + func the number of cycles is " + counter);
counter = 0;
results = new int[i+1];
func3(i);
System.out.println("For n="+i+" of func + func with remembering the number of cycles is " + counter);
}
}
public static int func(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return 2 * func(n - 1);
}
public static int func2(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return func2(n - 1) + func2(n - 1);
}
public static int func3(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
if (results[n] == 0){
results[n] = func3(n - 1) + func3(n - 1);
}
return results[n];
}
}
有这样的输出:
For n=0 of 2*func the number of cycles is 1
For n=0 of func + func the number of cycles is 1
For n=0 of func + func with remembering the number of cycles is 1
For n=1 of 2*func the number of cycles is 1
For n=1 of func + func the number of cycles is 1
For n=1 of func + func with remembering the number of cycles is 1
For n=2 of 2*func the number of cycles is 2
For n=2 of func + func the number of cycles is 3
For n=2 of func + func with remembering the number of cycles is 3
For n=3 of 2*func the number of cycles is 3
For n=3 of func + func the number of cycles is 7
For n=3 of func + func with remembering the number of cycles is 5
For n=4 of 2*func the number of cycles is 4
For n=4 of func + func the number of cycles is 15
For n=4 of func + func with remembering the number of cycles is 7
For n=5 of 2*func the number of cycles is 5
For n=5 of func + func the number of cycles is 31
For n=5 of func + func with remembering the number of cycles is 9
For n=6 of 2*func the number of cycles is 6
For n=6 of func + func the number of cycles is 63
For n=6 of func + func with remembering the number of cycles is 11
For n=7 of 2*func the number of cycles is 7
For n=7 of func + func the number of cycles is 127
For n=7 of func + func with remembering the number of cycles is 13
For n=8 of 2*func the number of cycles is 8
For n=8 of func + func the number of cycles is 255
For n=8 of func + func with remembering the number of cycles is 15
For n=9 of 2*func the number of cycles is 9
For n=9 of func + func the number of cycles is 511
For n=9 of func + func with remembering the number of cycles is 17
For n=10 of 2*func the number of cycles is 10
For n=10 of func + func the number of cycles is 1023
For n=10 of func + func with remembering the number of cycles is 19
For n=11 of 2*func the number of cycles is 11
For n=11 of func + func the number of cycles is 2047
For n=11 of func + func with remembering the number of cycles is 21
For n=12 of 2*func the number of cycles is 12
For n=12 of func + func the number of cycles is 4095
For n=12 of func + func with remembering the number of cycles is 23
For n=13 of 2*func the number of cycles is 13
For n=13 of func + func the number of cycles is 8191
For n=13 of func + func with remembering the number of cycles is 25
For n=14 of 2*func the number of cycles is 14
For n=14 of func + func the number of cycles is 16383
For n=14 of func + func with remembering the number of cycles is 27
For n=15 of 2*func the number of cycles is 15
For n=15 of func + func the number of cycles is 32767
For n=15 of func + func with remembering the number of cycles is 29
For n=16 of 2*func the number of cycles is 16
For n=16 of func + func the number of cycles is 65535
For n=16 of func + func with remembering the number of cycles is 31
For n=17 of 2*func the number of cycles is 17
For n=17 of func + func the number of cycles is 131071
For n=17 of func + func with remembering the number of cycles is 33
For n=18 of 2*func the number of cycles is 18
For n=18 of func + func the number of cycles is 262143
For n=18 of func + func with remembering the number of cycles is 35
For n=19 of 2*func the number of cycles is 19
For n=19 of func + func the number of cycles is 524287
For n=19 of func + func with remembering the number of cycles is 37
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