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)
{
if(pHead == NULL){
return NULL;
}
map<RandomListNode*,RandomListNode*>tmpMap;
RandomListNode* p1 = NULL;
RandomListNode* p2Head = NULL;
RandomListNode* tmpHead = pHead;
if(tmpHead){
p1 = new RandomListNode(tmpHead->label);
p2Head = p1;
tmpMap[tmpHead] = p1;
tmpHead = tmpHead->next;
}
while(tmpHead){
RandomListNode* tmp = new RandomListNode(tmpHead->label);
p1->next = tmp;
p1 = tmp;
tmpMap[tmpHead] = tmp;
tmpHead = tmpHead->next;
}
tmpHead = pHead;
map<RandomListNode*,RandomListNode*>::iterator it;
map<RandomListNode*,RandomListNode*>::iterator it_random;
while(tmpHead){
if(tmpHead->random){
it = tmpMap.find(tmpHead);
if(it != tmpMap.end()){
it_random = tmpMap.find(tmpHead->random);
if(it_random != tmpMap.end()){
it->second->random = it_random->second;
}
} }
tmpHead = tmpHead->next;
}
return p2Head;
}
};