S**I
发帖数: 15689 | 1 ☆─────────────────────────────────────☆
careerchange (Stupid) 于 (Sat Jun 4 20:25:58 2011, 美东) 提到:
void reversePrint(Node* head) {
if(head=NULL) {
cout << "Empty List. No node to print." << endl;
return;
}
if(head->next=NULL) {
cout << head->data << endl;
return;
} else {
Node* temp=head;
head=temp->next;
reversePrint(head);
cout << temp->data << endl;
return;
}
}
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speeddy (Wade) 于 (Sat Jun 4 20:30:33 2011, 美东) 提到:
.....
did not look into detail, but the first look find that
head = NULL...
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careerchange (Stupid) 于 (Sat Jun 4 20:49:01 2011, 美东) 提到:
You are right. Thanks. The old == versus = problem.
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careerchange (Stupid) 于 (Sat Jun 4 20:54:13 2011, 美东) 提到:
发现不定义新节点也没问题。
void reversePrint(Node* head) {
if(head==NULL) {
cout << "Empty List. No node to print." << endl;
return;
}
if(head->next==NULL) {
cout << head->data << endl;
return;
} else {
reversePrint(head->next);
cout << head->data << endl;
return;
}
}
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city (city) 于 (Sat Jun 4 20:54:44 2011, 美东) 提到:
你这Q+数字玩的是什么啊
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careerchange (Stupid) 于 (Sat Jun 4 21:51:43 2011, 美东) 提到:
开始是 Brainbench 的练习题。后来加的就是我自己想问的题目了，或者是面试题，或
者是我自己遇到的问题。
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RayCai (菜菜) 于 (Mon Jun 6 16:14:26 2011, 美东) 提到:
Only the last one will be printed.
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careerchange (Stupid) 于 (Sat Jun 11 11:59:06 2011, 美东) 提到:
Did you try it? I tried it. It printed all.
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timzheng (zmit) 于 (Sat Jun 11 15:06:08 2011, 美东) 提到:
"Empty List. No node to print"
That will always be printed out first.
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timzheng (zmit) 于 (Sat Jun 11 15:08:51 2011, 美东) 提到:
Actually no. This won't happen. But it's better to have this out of the
recursion.
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done (done) 于 (Sat Jun 11 15:22:00 2011, 美东) 提到:
你发现有什么错?感觉没什么错误.
但如果说以整个code来看,其实后面两个return是多余的,可删去,另外print的message
有bug,head是空就说是empty list,但因为用了递归,head->next为null时,不代表它是
empty list
手痒写了个,但没验证...
void reversePrint(Node * head){
if(!head){
cout << "the end ..." << endl; // print some message here if needed...
return;
}
reversePrint(head->next);
cout << head->data << endl;
}
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careerchange (Stupid) 于 (Sat Jun 11 17:18:21 2011, 美东) 提到:
use head==NULL, not head=NULL (this is always true).
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careerchange (Stupid) 于 (Sat Jun 11 20:15:06 2011, 美东) 提到:
你是对的。后面那两个return语句是多余的，因为程序会自然运行。
这个message应该没错。因为只有这个list为空时，这个message才会打印。否则第二个
return处理最终状态。
验证了一下，你是对的。并且你的处理终止状态比我的简单。
...
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careerchange (Stupid) 于 (Sat Jun 11 20:17:44 2011, 美东) 提到:
第二个return语句保证了message不会被打印，除非原来的list为空。
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done (done) 于 (Sat Jun 11 20:27:43 2011, 美东) 提到:
噢....谢谢,那你原来上面的return是有用的,我当时想的时候是把两个return去掉后才
觉得那个message有错.
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careerchange (Stupid) 于 (Sat Jun 11 20:47:22 2011, 美东) 提到:
完全不用return也是可以的。你的可以改成
void reversePrint(Node * head){
if(!head){
cout << "the end ..." << endl; // print some message here if needed...
} else {
reversePrint(head->next);
cout << head->data << endl;
}
}
我的可以改成
void reversePrint2(Node* head) {
if(head==NULL) {
cout << "Empty List. No node to print." << endl;
} else if(head->next==NULL) {
cout << head->data << " reversePrint2" << endl;
} else {
reversePrint2(head->next);
cout << head->data << " reversePrint2" << endl;
}
}
message
...
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speeddy (Wade) 于 (Sat Jun 11 21:14:04 2011, 美东) 提到:
Done's code is better.
void reversePrint(Node * head){
if(!head){
cout << "the end ..." << endl; // print some message here if needed..
.
} else {
reversePrint(head->next);
cout << head->data << endl;
}
}
Suppose you write this function and this function is very very long. Also, m
aybe there are a lot of #ifdef, #endif inside this function to support multi
ple platform. People look at this kind of code often get loss.
Then later on, another engineer jump in and modify this function by adding c
ode behind here:
} else {
reversePrint(head->next);
cout << head->data << endl;
}
/*****************/
add code here
/*****************/
}
But he did not pay attention to the code you write previously as the functio
n is so big. Normally, it should just return when the head is NULL. But now
it continue excuting without return.....Bug happens then.
...
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careerchange (Stupid) 于 (Sat Jun 11 21:26:46 2011, 美东) 提到:
I agree. It is better to use return than to use if else.
..
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lipingwu (ping) 于 (Sun Jun 12 14:19:09 2011, 美东) 提到:
It should work, but just is not very simple. We may be try following more
simple codes:
void reversePrint(Node *head){
if(head == NULL) return;
reversePrint(head->next);
cout << head->data << "\t";
}
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careerchange (Stupid) 于 (Sun Jun 12 15:18:45 2011, 美东) 提到:
It is the same as done's. |
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