剑指offer（22）-复杂链表的复制

题目描述

代码

/*
struct RandomListNode {
int label;
struct RandomListNode *next, *random;
RandomListNode(int x) :
label(x), next(NULL), random(NULL) {
}
};
*/
class Solution {
public:
/*
1、复制每个节点，如：复制节点A得到A1，将A1插入节点A后面
2、遍历链表，A1->random = A->random->next;
3、将链表拆分成原链表和复制后的链表
*/

RandomListNode* Clone(RandomListNode* pHead)
{
if(!pHead) return NULL;
RandomListNode *currNode = pHead;
while(currNode){
RandomListNode *node = new RandomListNode(currNode->label);
node->next = currNode->next;
currNode->next = node;
currNode = node->next;
}
currNode = pHead;
while(currNode){
RandomListNode *node = currNode->next;
if(currNode->random){
node->random = currNode->random->next;
}
currNode = node->next;
}
//拆分
RandomListNode *pCloneHead = pHead->next;
RandomListNode *tmp;
currNode = pHead;
while(currNode->next){
tmp = currNode->next;
currNode->next =tmp->next;
currNode = tmp;
}
return pCloneHead;
}
};