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)
C++ AC代码
struct RandomListNode {
int label;
struct RandomListNode *next, *random;
RandomListNode(int x) :
label(x), next(NULL), random(NULL) {
}
};
class Solution {
public:
RandomListNode* Clone(RandomListNode* pHead){
if(pHead == NULL){
return NULL;
}
RandomListNode* front = pHead;
RandomListNode* Head = new RandomListNode(-1);
RandomListNode* tmp = Head;
while(front){
RandomListNode* node = new RandomListNode(front->label);
tmp->next = node;
front = front->next;
tmp = tmp->next;
}
front = pHead;
tmp = Head->next;
while(front){
if(front->random){
tmp->random = this->Find(Head,front->random->label);
}
tmp = tmp->next;
front = front->next;
}
return Head->next;
}
RandomListNode* Find(RandomListNode* pHead, int k){
RandomListNode* front = pHead;
while(front){
if(front->label == k){
break;
}
front = front->next;
}
return front;
}
};