57. 删除链表中重复的结点

题目：在一个排序的链表中，如何删除重复的结点？例如，在图8.5(a)中重复结点被删除之后，链表如图8.5(b)所示。

思路：因为删除的有可能是头结点，所以采用二级指针，函数声明为 void delete(ListNode** pHead)；然后遍历链表，将当前结点的值和下一结点的值相同的结点都删除。为了保证删除之后链表仍然是相连的，序言保存当前即将删除结点的前一结点prevNode，和保存删除完后的下一个不重复结点lastNode，让prevNode指向lastNode。

#include<iostream>
#include<cstdio>
using namespace std;

struct ListNode
{
int                 m_nValue;
ListNode*           m_pNext;
};

ListNode* CreateListNode(int value)
{
ListNode* node = new ListNode();
node->m_nValue = value;
node->m_pNext = NULL;

return node;
}

void ConnectListNodes(ListNode* node1, ListNode* node2)
{
if (node1 != NULL && node2 != NULL)
{
node1->m_pNext = node2;
}
}

void PrintListNode(ListNode* pHead)
{
if (pHead == NULL)
{
return;
}

while (pHead != NULL)
{
cout << pHead->m_nValue << " ";
pHead = pHead->m_pNext;
}

cout << endl;

}

void DestroyList(ListNode* pHead)
{
ListNode* node = pHead;

while (node != NULL)
{
pHead = pHead->m_pNext;
delete node;
node = pHead;
}
}

void DeleteDuplication(ListNode** pHead)
{
if (pHead == NULL || *pHead == NULL)
{
return;
}

ListNode* prevNode = NULL;
ListNode* pNode = *pHead;

while (pNode != NULL)
{
ListNode* pNext = pNode->m_pNext;
bool needDelete = false;
if (pNext != NULL && pNext->m_nValue == pNode->m_nValue)
{
needDelete = true;
}

if (!needDelete)
{
prevNode = pNode;
pNode = pNode->m_pNext;
}
else
{
int value = pNode->m_nValue;
ListNode* toBeDel = pNode;
while (toBeDel != NULL && toBeDel->m_nValue == value)
{
pNext = toBeDel->m_pNext;
delete toBeDel;
toBeDel = NULL;

toBeDel = pNext;
}

if (prevNode == NULL)
{
*pHead = pNext;
}
else
{
prevNode->m_pNext = pNext;
}
pNode = pNext;
}
}
}

void test1()
{
ListNode* node1 = CreateListNode(1);
ListNode* node2 = CreateListNode(2);
ListNode* node3 = CreateListNode(3);
ListNode* node4 = CreateListNode(4);
ListNode* node5 = CreateListNode(5);
ListNode* node6 = CreateListNode(6);
ListNode* node7 = CreateListNode(5);

ConnectListNodes(node1, node2);
ConnectListNodes(node2, node3);
ConnectListNodes(node3, node4);
ConnectListNodes(node4, node5);
ConnectListNodes(node5, node7);
ConnectListNodes(node7, node6);

PrintListNode(node1);

DeleteDuplication(&node1);
PrintListNode(node1);

}

int main()
{
test1();

return 0;
}