Question

下面的代码将一个项目作为参数，并删除链接列表中所有项目的出现。它适用于我的测试。有什么我想念的吗？这段代码可以进一步改进吗？

void
LinkedList::DeleteAllOccurences(int key) {
   Node *temp = head;
   Node *prev = head;
   while(temp!=NULL) {
      if(temp->item == key){
         if(temp == head) {
            head = temp->next;
            delete temp;
            temp = head;
         } else {
            prev->next = temp->next;
            delete temp;
            temp = prev->next;
         }
      } else {
         prev = temp;
         temp = temp->next;
      }
   }
   return;
}

Answer 1

我相信你的代码有错误。删除head节点时，prev未正确更新（即），它仍将指向已删除的头节点。

我已经注释了您的代码并应用了修复[请原谅无偿的样式清理]：

void
LinkedList::DeleteAllOccurences(int key)
{
    Node *temp = head;
    Node *prev = head;

    while (temp != NULL) {
        if (temp->item == key) {
            // NOTE/BUG: after this, prev will _still_ be pointing to the
            // _deleted_ head node
            // NOTE/FIX: to fix this, prev must be set to the _updated_ head
            // node
            if (temp == head) {
                head = temp->next;

                // NOTE/FIX: add this:
#if 1
                prev = head;
#endif

                delete temp;
                temp = head;
            }
            else {
                prev->next = temp->next;
                delete temp;
                temp = prev->next;
            }
        }
        else {
            prev = temp;
            temp = temp->next;
        }
    }

    return;
}

可能也是另一个错误。并且，我认为有一种简化方法的方法。所以，为了比较：

void
LinkedList::DeleteAllOccurences(int key)
{
    Node *temp;
    Node *prev = NULL;
    Node *next;

    for (temp = head;  temp != NULL;  temp = next) {
        next = temp->next;

        if (temp->item != key) {
            prev = temp;
            continue;
        }

        if (prev != NULL)
            prev->next = next;
        else
            head = next;

        delete temp;
    }
}

Answer 2

另一种方法是使用std::list<>并让它完成所有繁重的工作。以下是一个适合您需要的示例实现：

#include <list>
#include <iostream>

using namespace std;

int main()
{
    // Create a list.
    list<int> myList;

    // Add some numbers: 2, 3, 2, and 5.
    myList.push_back(2);
    myList.push_back(3);
    myList.push_back(2);
    myList.push_back(5);

    // Print the contents of the list.
    // Will output 2, 3, 2, 5.
    for(auto item : myList)
        cout << item << endl;

    // Remove all numbers equal to 2.
    myList.remove(2);

    // Print the contents of the list.
    // Will output 3 and 5.
    for(auto item : myList)
        cout << item << endl;

    return 0;
}

删除链接列表中所有项目的出现

2 个答案: