6.逆值/翻轉單鏈表
思想同從頭打印鏈表一樣,但應注意在反向連接鏈表時的指向問題
void ListTranspose(ListNode **ppList,ListNode *pList)//轉置單鏈表
{
assert(ppList);
if((*ppList == NULL)||((*ppList)->next == NULL))
return;
//當鏈表沒有節點和只有一個節點的時候,轉置不改變
if(pList->next != NULL) //
ListTranspose(ppList,pList->next);
else
{
*ppList = pList;//找到最後一個節點,並用頭節點指向
return;
}
(pList->next)->next = pList; //
pList->next = NULL;
}
7. 單鏈表排序(冒泡排序)
冒泡排序是兩兩比較,N個數,需要N-1趟,每一趟可以找出最大的數放在這一趟的最右邊:
void ListBubb(ListNode **ppList)//冒泡排序
{
assert(ppList);
if(*ppList == NULL)//鏈表爲空時,直接返回
return;
ListNode *head = *ppList;//每次一趟從頭節點開始
ListNode *tail = NULL; //用於結束標誌,每一趟後tail向前一個節點挪
while((*ppList)->next != tail)
{
head = *ppList;
while(head->next != tail)
{
if((head->data) > (head->next->data))
{
DataType tmp = head->data;
head->data = head->next->data;
head->next->data = tmp;
}
head = head->next;
}
tail = head;
}
}
8.合併兩個有序鏈表,合併後依然有序
最簡單的方法就是用第一個鏈表的尾部連上另一個鏈表頭節點,再用冒泡排序:
ListNode* ListLink(ListNode **ppList_1,ListNode **ppList_2)
{
assert(ppList_1&&ppList_2);
ListNode *tail = *ppList_1;
while(tail->next != NULL)
{
tail = tail->next;
}
tail->next = *ppList_2;
ListBubb(ppList_1);
return *ppList_1;
}
但這種方法時間複雜太大,有沒有更好的辦法呢?
先找兩個鏈表第一個節點的小的頭鏈表,然後在這鏈表後面插兩兩比較小的節點。直至其中一個鏈表遇到NULL;
//鏈接鏈表並排序
ListNode* ListLink(ListNode **ppList_1,ListNode **ppList_2)
{
assert(ppList_1&&ppList_2);
ListNode *pList_1 = *ppList_1;
ListNode *pList_2 = *ppList_2;
if(pList_1->data > pList_2->data)//找到頭節點小的鏈表
{
*ppList_1 = *ppList_2;
pList_2 = pList_2->next;
}
else
{
pList_1 = pList_1->next;
}
ListNode *pList = *ppList_1;
while(pList_1 && pList_2)
{
if(pList_1->data > pList_2->data)
{
pList->next = pList_2;
pList_2 = pList_2->next;
}
else
{
pList->next = pList_1;
pList_1 = pList_1->next;
}
pList = pList->next;
}
if(pList_1)//遇到NULL
{
pList = pList_2;
}
else
{
pList = pList_1;
}
return *ppList_1;
}