两个链表的并集 - C++
任何帮助都会有所帮助。我编写了一段代码来查找两个链表的并集。但是，我在某一部分遇到了无限循环。我在代码中指出了。请帮我找出错误。谢谢。
//Finds the union of two linked lists.
nodeType* unionLL(nodeType *&headA, nodeType *&headB)
{
nodeType *tempA, *tempB, *newNode;
tempA = headA;
tempB = headB;
bool match = false;
while (tempB -> link != NULL)
{
while (tempA -> link != NULL)
{
outfile <<"The infinite loop occurs here " << endl;
if (tempB -> intVal == tempA -> intVal)
{
match = true;
}
tempA = tempA -> link;
}
if (!match)
{
newNode = new nodeType;
newNode -> intVal = tempB -> intVal;
newNode -> link = NULL;
tempA -> link = newNode;
}
tempA = headB;
tempB = tempB -> link;
}
return headB;
}
Any assistance will be helpful. I have written a code to find the union of two linked list. However, I am getting an infinite loop at one section. I indicate it in the code. Please help me identify the error. Thanks.
//Finds the union of two linked lists.
nodeType* unionLL(nodeType *&headA, nodeType *&headB)
{
nodeType *tempA, *tempB, *newNode;
tempA = headA;
tempB = headB;
bool match = false;
while (tempB -> link != NULL)
{
while (tempA -> link != NULL)
{
outfile <<"The infinite loop occurs here " << endl;
if (tempB -> intVal == tempA -> intVal)
{
match = true;
}
tempA = tempA -> link;
}
if (!match)
{
newNode = new nodeType;
newNode -> intVal = tempB -> intVal;
newNode -> link = NULL;
tempA -> link = newNode;
}
tempA = headB;
tempB = tempB -> link;
}
return headB;
}
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您还没有确定链接列表是否已排序 - 所以我们应该假设没有。您尚未确定可以修改哪些列表；从表面上看，两个列表都可以由该函数修改。您返回
headB，这表明结果应该是可从headB访问的结果集应包含该列表中的每个元素，以及可从可访问的每个元素的新元素>headA尚未通过headB找到。那么，从表面上看，您的伪代码应该是：
保持 listA 不变。您的代码没有实现此逻辑。
You have not identified whether the linked lists are sorted or not - so we should assume not. You have not identified which list can be modified; superficially, both lists could be modified by the function. You return
headBwhich suggests that the result should be that the result set accessible fromheadBshould contain every element in that list, plus a new element for each element accessible fromheadAthat is not already found viaheadB.Superficially, then, your pseudo-code should be:
leaving listA untouched. Your code does not implement this logic.
我认为你没有检查它是否在 tempA 循环中匹配（在 B 中也没有）
I think you are not checking if it matches in tempA loop (neither in B's)