题目描述:
输入一个复杂链表(每个节点中有节点值,以及两个指针,一个指向下一个节点,另一个特殊指针指向任意一个节点),返回结果为复制后复杂链表的head。
注意:输出结果中请不要返回参数中的节点引用,否则判题程序会直接返回空。
/* struct RandomListNode { int label; struct RandomListNode *next, *random; RandomListNode(int x) : label(x), next(NULL), random(NULL) { } }; */
class Solution {
public:
RandomListNode* Clone(RandomListNode* pHead)
{
CloneNodes(pHead);
ConnectRandom(pHead);
return ReConnectNodes(pHead);
}
private:
void CloneNodes(RandomListNode* pHead){
RandomListNode *pNode = pHead;
while(pNode != NULL){
RandomListNode *pCloned = new RandomListNode(0);
pCloned->label = pNode->label;
pCloned->next = pNode->next;
pCloned->random = NULL;
pNode->next = pCloned;
pNode = pCloned->next;
}
}
void ConnectRandom(RandomListNode* pHead){
RandomListNode *pNode = pHead;
while(pNode){
RandomListNode *pCloned = pNode->next;
if(pNode->random != NULL)
pCloned->random = pNode->random->next;
pNode = pCloned->next;
}
}
RandomListNode* ReConnectNodes(RandomListNode* pHead){
RandomListNode *pNode = pHead;
RandomListNode *pClonedNode = NULL;
RandomListNode *pClonedHead = NULL;
if(pNode != NULL){
pClonedNode = pClonedHead = pNode->next;
pNode->next = pClonedNode->next;
pNode = pNode->next;
}
while(pNode){
pClonedNode->next = pNode->next;
pClonedNode = pClonedNode->next;
pNode->next = pClonedNode->next;
pNode = pNode->next;
}
return pClonedHead;
}
};
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