Question

这是我要问的同一个问题： deleting even nodes from a doubly link list c++

不同之处在于，我想了解我的代码有什么问题。我不想只是接受一种完全不同的方法而不理解为什么我的代码不起作用。以下是我的代码的两个版本我想知道这两个版本有什么问题。它们都给我分段错误。

int removeEven(node *&head)
{
if(!head)               //Base case: end of list reached
    return 0;
int count = removeEven(head->next);     //Recurse to end of list
if(head->data % 2 != 0)
    return count;
else{
    ++count;
    if(head->next){
        head->next->previous = head->previous;
    }
    if(head->previous){
        head->previous->next = head->next;
    } if(!(head->previous)){
        node* temp = head;
        head = head->next;
        delete temp;
    }
    else
        delete head;
}
return count;
}

第二个将count = 0作为默认arg。

int removeEven(node *&head, int count)
if(head && head->data % 2 != 0) //not null, not even
{
    removeEven(head->next, count);
}
else if(head != NULL){      //not null, yes even
   ++count;
    if(head->next)
        head->next->previous = head->previous;
    if(head->previous)
        head->previous->next = head->next;
    node* temp = head;
    head = head->next;
    delete temp;
   removeEven(head, count);
}
    return count;       //base case: null
}

Answer 1

int removeEven(node *&head)
{
  if(!head)               //Base case: end of list reached
    return 0;
  int count = removeEven(head->next);     //Recurse to end of list
  if(head->data % 2 != 0)
    return count;
  else{
    ++count;
    // CORRECT WAY!!! copy the old pointer in a temp
    node *t = head;
    if(head->next){
      head->next->previous = head->previous;
    }
    if(head->previous){
      // WARNING!!! here you are ACTUALLY modifying head to nullptr
      head->previous->next = head->next;
    }
    // CORRECT WAY!!! delete the temp pointer
    delete t;
    // WARNING!!! here you are trying to access a nullptr in head
    // if(!(head->previous)){
    //   node* temp = head;
    //   head = head->next;
    //   delete temp;
    // }
    // else
    //   delete head;
  }
  return count;
}

我已经了解了valgrind和gdb

的根本原因

valgrind ./a.out 
==2729== Memcheck, a memory error detector
==2729== Copyright (C) 2002-2013, and GNU GPL'd, by Julian Seward et al.
==2729== Using Valgrind-3.10.0 and LibVEX; rerun with -h for copyright info
==2729== Command: ./a.out
==2729== 
[9] [8] [7] [6] [5] [4] [3] [2] [1] [0] 
==2729== Invalid read of size 8
==2729==    at 0x4008F5: removeEven(node*&) (list1.cpp:26)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x400AFD: main (list1.cpp:81)
==2729==  Address 0x10 is not stack'd, malloc'd or (recently) free'd
==2729== 
==2729== 
==2729== Process terminating with default action of signal 11 (SIGSEGV)
==2729==  Access not within mapped region at address 0x10
==2729==    at 0x4008F5: removeEven(node*&) (list1.cpp:26)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x40087A: removeEven(node*&) (list1.cpp:15)
==2729==    by 0x400AFD: main (list1.cpp:81)
==2729==  If you believe this happened as a result of a stack
==2729==  overflow in your program's main thread (unlikely but
==2729==  possible), you can try to increase the size of the
==2729==  main thread stack using the --main-stacksize= flag.
==2729==  The main thread stack size used in this run was 8720384.
==2729== 
==2729== HEAP SUMMARY:
==2729==     in use at exit: 240 bytes in 10 blocks
==2729==   total heap usage: 10 allocs, 0 frees, 240 bytes allocated
==2729== 
==2729== LEAK SUMMARY:
==2729==    definitely lost: 24 bytes in 1 blocks
==2729==    indirectly lost: 0 bytes in 0 blocks
==2729==      possibly lost: 0 bytes in 0 blocks
==2729==    still reachable: 216 bytes in 9 blocks
==2729==         suppressed: 0 bytes in 0 blocks
==2729== Rerun with --leak-check=full to see details of leaked memory
==2729== 
==2729== For counts of detected and suppressed errors, rerun with: -v
==2729== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0)

Compilation segmentation fault at Fri Jul 28 09:46:51

Answer 2

我使用以偶数项开头的列表测试了您的代码，但它失败了。教授提供的所有测试用例都以偶数项开始。这是解决方案：

int removeEven(node *&head)
{
if(!head)               //Base case: end of list reached
    return 0;
int count = removeEven(head->next);     //Recurse to end of list
if(head->data % 2 != 0)
    return count;
else{
    ++count;
    node *t = head;
    if(head->next)
        head->next->previous = head->previous;
    if(head->previous)
        head->previous->next = head->next;
    if(head && !head->previous)
        head = head->next;
    delete t;
}
return count;
}

问题是如果头节点被删除，实际的原始头指针会丢失。现在我检查一下我们是否处于实际头部，因为head-＆gt; previous应为NULL，然后我将实际的头指针设置为列表中的下一个节点，然后再删除。

从双向链表C ++中递归删除偶数项

2 个答案: