我为类编写了一个赋值，我必须编写一个递归函数来使用递归输出一组数字，非常简单，我设法做到了。现在我必须编写另一个函数，使用递归以相反的顺序输出相同的数字，这就是我被卡住的地方，我对如何使用递归打印出反向输出感到茫然。所以我的问题仍然存在，如何使用递归打印出我的第一个函数的反向输出？我应该颠倒输出的顺序，而不是反过来计算序列。
另外，另一个问题，为什么我的计数器不工作？谢谢你的帮助。
代码：
#include <iostream>
using namespace std;
void reverseOJP(int refD, int D) {
// ?
}
void lengthOJP (int D, int count) {
cout << endl << "The length of the OJP for " << D << " is " << count << endl;
}
int OJP(int D, int count) {
count++;
if (D == 1) { // Base case
return count;
} else if ((D % 2) != 0) { // odd numbers
D = ((D * 3) + 1); // if D is odd
cout << D << " ";
OJP(D, count); // Recursive Call
} else { // even
D /= 2;
cout << D << " ";
OJP(D, count);
}
return count;
}
int main() {
int D, count = 0; // variables
int refD;
cout << "Positive integer: " << endl;
D = 12;
// OJP output
cout << "The OJP for " << D << endl;
cout << D << " ";
refD = D; // D reference for output purposes
//OJP(D, count, refD); // OJP call
count = count + OJP (D, count);
cout << "The OJP for " << D << endl;
cout << "1 ";
reverseOJP(refD, D); // call and print reverse OJP
lengthOJP(refD, count); // Once reverse is printed print length
return 0;
}
期望的输出：
Enter a positive integer: 12
The OJP for 12: 12 6 3 10 5 16 8 4 2 1
The reverse OJP for 12: 1 2 4 8 16 5 10 3 6 12
The length of the OJP for 12 is 10
答案 0 :(得分：0)
要反转输出，只需重新排序int OJP(int D, int count)：
打印当前号码，而不是之前打印 递归调用（即返回指令之前）：
void reverseOJP(int D, int refD) {
if (D == 1) { // Base case
return ;
} else if ((D % 2) != 0) { // odd numbers
D = ((D * 3) + 1); // if D is odd
reverseOJP(D, D); // Recursive Call
cout << refD << " ";
} else { // even
D /= 2;
reverseOJP(D, D);
cout << refD << " ";
}
}
列出的代码只是稍微改编自OJP的复制粘贴，尚未优化。
答案 1 :(得分：0)
至于返回的计数，请查看下面的示例输出，看看是否可以找到与...＃count; count＆＃34;在main（）比你预期的。当你可以看到代码正在做什么时，更容易解决这类问题，我喜欢在我的代码中添加一些调试打印，如下所示：
#include <iostream>
#include <string> // added for debug stuff.
using namespace std;
using namespace std;
void reverseOJP(int refD, int D) {
// ?
}
void lengthOJP (int D, int count) {
cout << endl << "The length of the OJP for " << D << " is " << count << endl;
}
//***** debug stuff ******
int debug_depth = 0; // caller must increment & decrement when entering & leaving funct.
void debug_indent( ) {
for( int i = 0; i < debug_depth; ++i ) cout << ": ";
}
void debug_enter( const string & funct_name ) {
++debug_depth;
debug_indent(); cout << ">" << funct_name << "()" << endl;
}
void debug_exit( const string & funct_name ) {
--debug_depth;
debug_indent(); cout << "<" << funct_name << "()" << endl;
}
//***** end stuff ******
int OJP(int D, int count) {
// debug output convention, ">" going into a fnct and "<" leaving it.
// doing the simple indent makes it easier to see what fnct is doing.
debug_enter("OJP");
debug_indent(); cout << "OJP(): my count=" << count << endl;
count++;
debug_indent(); cout << "OJP(): after ++, now count=" << count << endl;
if (D == 1) { // Base case
debug_indent(); cout << "debug.OJP(): D==1, returning count=" << count << endl;
debug_exit("OJP");
return count;
} else if ((D % 2) != 0) { // odd numbers
D = ((D * 3) + 1); // if D is odd
// SAVE: cout << D << " ";
debug_indent(); cout << "odd, now D=" << D << endl;
OJP(D, count); // Recursive Call
} else { // even
D /= 2;
// SAVE: cout << D << " ";
debug_indent(); cout << "even, now D=" << D << endl;
OJP(D, count);
}
debug_indent(); cout << "debug.OJP(): returning count=" << count << endl;
debug_indent(); cout << "debug.OJP(): returning count=" << count << endl;
debug_exit("OJP");
// *** When you return count here, what is the value going back to main() ?
return count;
}
int main() {
int D, count = 0; // variables
int refD;
cout << "Positive integer: " << endl;
D = 12;
// OJP output
cout << "The OJP for " << D << endl;
cout << D << " ";
refD = D; // D reference for output purposes
//OJP(D, count, refD); // OJP call
cout << "main(): before OJP(), count=" << count << endl;
count = count + OJP (D, count);
cout << "main(): now, count=" << count << endl;
cout << "The OJP for " << D << endl;
cout << "1 ";
reverseOJP(refD, D); // call and print reverse OJP
lengthOJP(refD, count); // Once reverse is printed print length
return 0;
}
示例输出：
Positive integer:
The OJP for 12
12 debug.main(): before OJP(), count=0
: >OJP()
: OJP(): my count=0
: OJP(): after ++, now count=1
: even, now D=6
: : >OJP()
: : OJP(): my count=1
: : OJP(): after ++, now count=2
: : even, now D=3
: : : >OJP()
: : : OJP(): my count=2
: : : OJP(): after ++, now count=3
: : : odd, now D=10
: : : : >OJP()
: : : : OJP(): my count=3
: : : : OJP(): after ++, now count=4
: : : : even, now D=5
: : : : : >OJP()
: : : : : OJP(): my count=4
: : : : : OJP(): after ++, now count=5
: : : : : odd, now D=16
: : : : : : >OJP()
: : : : : : OJP(): my count=5
: : : : : : OJP(): after ++, now count=6
: : : : : : even, now D=8
: : : : : : : >OJP()
: : : : : : : OJP(): my count=6
: : : : : : : OJP(): after ++, now count=7
: : : : : : : even, now D=4
: : : : : : : : >OJP()
: : : : : : : : OJP(): my count=7
: : : : : : : : OJP(): after ++, now count=8
: : : : : : : : even, now D=2
: : : : : : : : : >OJP()
: : : : : : : : : OJP(): my count=8
: : : : : : : : : OJP(): after ++, now count=9
: : : : : : : : : even, now D=1
: : : : : : : : : : >OJP()
: : : : : : : : : : OJP(): my count=9
: : : : : : : : : : OJP(): after ++, now count=10
: : : : : : : : : <OJP()
: : : : : : : : : debug.OJP(): returning count=9
: : : : : : : : <OJP()
: : : : : : : : debug.OJP(): returning count=8
: : : : : : : <OJP()
: : : : : : : debug.OJP(): returning count=7
: : : : : : <OJP()
: : : : : : debug.OJP(): returning count=6
: : : : : <OJP()
: : : : : debug.OJP(): returning count=5
: : : : <OJP()
: : : : debug.OJP(): returning count=4
: : : <OJP()
: : : debug.OJP(): returning count=3
: : <OJP()
: : debug.OJP(): returning count=2
: <OJP()
: debug.OJP(): returning count=1
<OJP()
debug.main(): now, count=1
The OJP for 12
1
The length of the OJP for 12 is 1