{

int m_nValue;

ListNode* m_pNext;

{

if(pListHead) {

if(*pListHead) {

if(*pListHead==pToBeDeleted) { //如果这个节点在头部

/*if(pToBeDeleted->m_pNext==NULL) { //如果这个节点唯一的

//可以简化代码

*pListHead = pToBeDeleted->m_pNext;

delete pToBeDeleted;

} else if(pToBeDeleted->m_pNext==NULL) { //如果此节点在尾部，只能从头遍历

ListNode* prev = find_prev(*pListHead, pToBeDeleted);

prev->m_pNext = NULL;

delete pToBeDeleted;

} else { //只剩下在中间部分的情况了

ListNode* pNext = pToBeDeleted->m_pNext;

ListNode* pNextNext = pNext->m_pNext;

pToBeDeleted->m_nValue = pNext->m_nValue;

pToBeDeleted->m_pNext = pNextNext;

delete pNext;

}

}

}

{

if(pHead) {

ListNode* head = pHead;

cout << head->m_nValue << ' ';

while(head->m_pNext) {

head = head->m_pNext;

cout << head->m_nValue << ' ';

}

cout << endl;

}

{

ListNode* root = NULL;

ListNode* node6 = add_to_head(&root, 6);

ListNode* node5 = add_to_head(&root, 5);

ListNode* node4 = add_to_head(&root, 4);

ListNode* node3 = add_to_head(&root, 3);

ListNode* node2 = add_to_head(&root, 2);

ListNode* node1 = add_to_head(&root, 1);

print_list_head_to_tail(root);

DeleteNode(&root, node3);

print_list_head_to_tail(root);

//由于为了实现 O(1) 的删除效率，所以删除node5的时候其实node6已经不存在了

//所以执行删除node6才会出错

DeleteNode(&root, node5);

//DeleteNode(&root, node6);替换为再次删除node5才可以，因为此时它在尾部

DeleteNode(&root, node5);

print_list_head_to_tail(root);

DeleteNode(&root, node1);

print_list_head_to_tail(root);

DeleteNode(&root, node2);

DeleteNode(&root, node3);

print_list_head_to_tail(root);

return 0;