链表
文章目录
c
struct List
{
int data;
struct List* next;
}
创建链表
单链表
实现
c
struct List* listCreate()
{
int data;
struct List* head = NULL;
struct List* pre = NULL;
struct List* current = NULL;
while(scanf("%d",&data) && data != -1)
{
current = (struct List*)malloc(sizeof(struct List));
if(head == NULL)
head = current;
else
pre->next = current;
current->next = NULL;
current->data = data;
pre = current;
}
return head;
}
错例
c
struct List* listCreate()
{
int data;;
struct List* current = NULL;
struct List* head = current;
while (scanf("%d", &data) && data != -1)
{
current = (struct List*)malloc(sizeof(struct List));
if (head == NULL)
head = current;
current->data = data;
current = current->next;
}
return head;
}
在使用malloc函数开辟的空间中,不要进行指针的移动,因为一旦移动之后可能出现申请的空间和释放空间大小的不匹配
循环链表
单独创建
c
struct List* circle_listCreate()
{
int data;
struct List* head = NULL;
struct List* pre = NULL;
struct List* current = NULL;
while(scanf("%d",&data) && data != -1)
{
current = (struct List*)malloc(sizeof(struct List));
if(head == NULL)
head = current;
else
pre->next = current;
current->next = head;
current->data = data;
pre = current;
}
return head;
}
逐节点创建
c
void Append(struct List** L,int data)
{
struct List* head = *L;
struct List* newNode = NULL;
if((*L) == NULL)
{
(*L) = (struct List*)malloc(sizeof(struct List));
(*L)->data = data;
head = (*L);
(*L)->next = head;
}
else
{
while ((*L)->next != head){
(*L) = (*L)->next;
}
newNode = (struct List*)malloc(sizeof(struct List));
newNode->data = data;
(*L)->next = newNode;
newNode->next = head;
*L = head;
}
}
约瑟夫环问题
c
void Append(struct List** L,int data)
{
struct List* head = *L;
struct List* newNode = NULL;
if((*L) == NULL)
{
(*L) = (struct List*)malloc(sizeof(struct List));
(*L)->data = data;
head = (*L);
(*L)->next = head;
}
else
{
while ((*L)->next != head){
(*L) = (*L)->next;
}
newNode = (struct List*)malloc(sizeof(struct List));
newNode->data = data;
(*L)->next = newNode;
newNode->next = head;
*L = head;
}
}
void Display(struct List* L,int num)
{
struct List* head = L;
struct List* pre = NULL;
struct List* kill = NULL;
int nodeNum = 0;
while (L->next != head)
{
nodeNum++;
L = L->next;
}
pre = L;
L = L->next;
nodeNum++;
while (nodeNUm)
{
if (nodeNum == 1)
{
printf("%d",L->data);
free(L);
return;
}
for (int i=1; i < m; i++)
{
pre = L;
L = L->next;
}
printf("%d ", L->data);
kill = L;
L = L->next;
free(kill);
nodeNum--;
}
}
删除节点
实现方式一:
c
struct list* listDelete(struct list* L,int data)
{
struct list* pre = L;
struct list* head = L;
struct list* kill;
while(head != NULL && head->data == m)
{
kill = head;
head = head->next;
free(kill);
}
if(head == NULL)
return head;
pre = head;
kill = head->next;
while(kill!=NULL)
{
if(kill->data == m)
{
pre->next = kill->next;
free(kill);
kill = pre->next;
}
else
{
pre = kill;
kill = kill->next;
}
}
return head;
}
实现方式二:
c
struct list* listDelete(struct list** L,int data)
{
struct list* head = (*L), * pre = (*L);
struct list* newL = *L;
struct list* kill = NULL;
while (*L !- NULL)
{
if((*L)->data == data)
{
if((*L) == newL)
newL == newL->next;
else
pre->next = (*L)->next;
kill = (*L);
(*L) = (*L)->next;
free(kill);
}
else
{
pre = (*L);
(*L) = (*L)->next;
}
}
*L = newL;
return head;
}
删除节点并建立新链表
c
struct list* list_Delete_Create(struct list** L) //数据为奇数存为新链表
{
struct list* newhead = NULL, * newcurrent = NULL, * newpre = NULL;
struct list* newL = *L;
struct list* kill = NULL;
struct list* pre = *L;
while (*L)
{
if((*L)->data%2 == 1)
{
newcurrent = (struct list*)malloc(sizeof(struct list));
if(newhead == NULL)
newhead = newcurrent;
else
newpre->next = newcurrent;
newcurrent->data = (*L)->data;
newcurrent->next = NULL;
newpre = newcurrent;
if((*L) == newL)
newL = newL->next;
else
pre-next = (*L)->next;
kill = (*L);
(*L)=(*L)->next;
free(kill);
}
else
{
pre = (*L);
(*L) = (*L)->next;
}
}
*L = newL;
return newhead;
}
逆置链表
实现:
c
struct list* reverse(struct list* L)
{
struct list* newhead = NULL, * current;
while (L != NULL)
{
current = (struct list*)malloc(sizeof(struct list));
current->data = L->data;
L = L->next;
current->next = newhead;
newhead = current;
}
free(L);
return newhead;
}