时间:2022-06-03 11:09:26 | 栏目:C代码 | 点击:次
图解

typedef int DataType;
typedef struct DListNode
{
DataType data;
DListNode* prev;
DListNode* next;
}DListNode;
void DListInint(DListNode** pphead)
{
*pphead = (DListNode*)malloc(sizeof(DListNode));
(*pphead)->next = (*pphead);
(*pphead)->prev = (*pphead);
}
或者使用返回节点的方法也能实现初始化
DListNode* DListInit()
{
DListNode* phead = (DListNode*)malloc(sizeof(DListNode));
phead->next = phead;
phead->prev = phead;
return phead;
}
DListNode* BuyDListNode(DataType x)
{
DListNode* temp = (DListNode*)malloc(sizeof(DListNode));
if (temp == NULL)
{
printf("malloc fail\n");
exit(-1);
}
temp->prev = NULL;
temp->next = NULL;
temp->data = x;
return temp;
}
void DListPushBack(DListNode* phead,DataType x)
{
DListNode* newnode = BuyDListNode(x);
DListNode* tail = phead->prev;
tail->next = newnode;
newnode->prev = tail;
newnode->next = phead;
phead->prev = newnode;
}
void DListNodePrint(DListNode* phead)
{
DListNode* cur = phead->next;
while (cur != phead)
{
printf("%d->", cur->data);
cur = cur->next;
}
printf("NULL\n");
}
void DListNodePushFront(DListNode* phead, DataType x)
{
DListNode* next = phead->next;
DListNode* newnode = BuyDListNode(x);
next->prev = newnode;
newnode->next = next;
newnode->prev = phead;
phead->next = newnode;
}
void DListNodePopBack(DListNode* phead)
{
if (phead->next == phead)
{
return;
}
DListNode* tail = phead->prev;
DListNode* prev = tail->prev;
prev->next = phead;
phead->prev = prev;
free(tail);
tail = NULL;
}
void DListNodePopFront(DListNode* phead)
{
if (phead->next == phead)
{
return;
}
DListNode* firstnode = phead->next;
DListNode* secondnode = firstnode->next;
secondnode->prev = phead;
phead->next = secondnode;
free(firstnode);
firstnode = NULL;
}
DListNode* DListNodeFind(DListNode* phead, DataType x)
{
DListNode* firstnode = phead->next;
while (firstnode != phead)
{
if (firstnode->data == x)
{
return firstnode;
}
firstnode = firstnode->next;
}
return NULL;
}
void DListNodeInsert(DListNode* pos, DataType x)
{
DListNode* prev = pos->prev;
DListNode* newnode = BuyDListNode(x);
newnode->next = pos;
newnode->prev = prev;
prev->next = newnode;
pos->prev = newnode;
}
void DListNodeErase(DListNode* pos)
{
DListNode* prev = pos->prev;
DListNode* next = pos->next;
prev->next = next;
next->prev = prev;
free(pos);
pos = NULL;
}
链表的中间节点(力扣)
给定一个头结点为 head 的非空单链表,返回链表的中间结点。
如果有两个中间结点,则返回第二个中间结点。
输入:[1,2,3,4,5]
输出:此列表中的结点 3 (序列化形式:[3,4,5])
返回的结点值为 3 。 (测评系统对该结点序列化表述是 [3,4,5])。
注意,我们返回了一个 ListNode 类型的对象 ans,
这样:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, 以及 ans.next.next.next = NULL.
来源:力扣(LeetCode)
思路:快慢指针
取两个指针,初始时均指向head,一个为快指针(fast)一次走两步,另一个为慢指针(slow)一次走一步,当快指针满足fast==NULL(偶数个节点)或者fast->next==NULL(奇数个节点)时,slow指向中间节点,返回slow即可。
struct ListNode* middleNode(struct ListNode* head)
{
struct ListNode* fast=head;
struct ListNode* slow=head;
while(fast&&fast->next)
{
fast=fast->next->next;
slow=slow->next;
}
return slow;
}