一个链表的每个节点,有一个指向next指针指向下一个节点,还有一个rand指针指向这个链表中的一个随机节点或NULL,现在要求复制一个单链表来实现这个链表,返回复制后的新链表。

来源:互联网 发布:dnf安图恩数据异常 编辑:程序博客网 时间:2024/05/18 06:22

假设有List 为:  1    3      5     7      9

List* phead为此链表的头结点


typedef struct ComNode
{
int _data;
struct ComNode* next;
struct ComNode* rand;
}ComNode;


ComNode* BuyNode(Data x)
{
ComNode *phead_1;
phead_1= (ComNode*)malloc(sizeof(ComNode));
phead_1->_data = x;
phead_1->next = NULL;
phead_1->rand= NULL;
return phead_1;
}
ComNode* InitComNode(List** pphead)
{
int count = 1;
int count_2 = 0;
List* start = *pphead;
ComNode* phead = BuyNode((*pphead)->arrary);
ComNode* start_1 = phead;
ComNode* cur = phead;;
int count_1 = rand() % 2;
if (count_1 == 1)
{
start_1->rand = phead;
}
else
{
start_1->rand = NULL;
}
while ((start)->next!=NULL)
{
start = start->next;
start_1->next = BuyNode(start->arrary);
start_1 = start_1->next;
count++;
count_2 = rand() % count;
while (count_2 > 0)
{
cur = cur->next;
count_2--;
}
start_1->rand = cur;
cur = phead;
}
return phead;
}

void PrintCom_1(ComNode* phead)
{
ComNode* start = phead;
while (start)
{
printf("%d ", start->_data);
start = start->next;
}
printf("\n");
}


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