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)
解法一
首先从链表节点的next开始复制,先获得一条单向链表,在获得单向链表的过程中,使用map来将原链表节点和复制后得到的链表节点建立映射; 接着复制链表节点中的特殊指针,根据原链表中的特殊指针在map中获取对应的复制节点;
解法一代码
/*
struct RandomListNode {
int label;
struct RandomListNode *next, *random;
RandomListNode(int x) :
label(x), next(NULL), random(NULL) {
}
};
*/
class Solution {
public:
map<RandomListNode*, RandomListNode*> RandomMap;
RandomListNode* Clone(RandomListNode* pHead)
{
if(pHead == nullptr) return nullptr;
RandomListNode *cHead = new RandomListNode(-1);
CopyLinkForNext(pHead,cHead);
CopyLinkForRandom(pHead,cHead);
return cHead;
}
void CopyLinkForNext(RandomListNode* pHead, RandomListNode *cHead) {
if (cHead == nullptr) return;
RandomListNode* chead = cHead;
RandomListNode* phead = pHead;
while (phead != nullptr) {
chead->label = phead->label;
RandomMap[phead] = chead;
phead = phead->next;
if (phead == nullptr) break;
RandomListNode* temp = new RandomListNode(-1);
chead->next = temp;
chead = chead->next;
}
}
void CopyLinkForRandom(RandomListNode *pHead, RandomListNode* cHead) {
if (cHead == nullptr) return;
RandomListNode* chead = cHead;
RandomListNode* phead = pHead;
while (phead != nullptr) {
if (phead->random == nullptr) {
chead = chead->next;
phead = phead->next;
continue;
}
chead->random = RandomMap[phead->random];
chead = chead->next;
phead = phead->next;
}
}
};
解法二
解法二代码
public class Solution {
public RandomListNode Clone(RandomListNode pHead) {
if(pHead == null) {
return null;
}
RandomListNode currentNode = pHead;
//1、复制每个结点,如复制结点A得到A1,将结点A1插到结点A后面;
while(currentNode != null){
RandomListNode cloneNode = new RandomListNode(currentNode.label);
RandomListNode nextNode = currentNode.next;
currentNode.next = cloneNode;
cloneNode.next = nextNode;
currentNode = nextNode;
}
currentNode = pHead;
//2、重新遍历链表,复制老结点的随机指针给新结点,如A1.random = A.random.next;
while(currentNode != null) {
currentNode.next.random = currentNode.random==null?null:currentNode.random.next;
currentNode = currentNode.next.next;
}
//3、拆分链表,将链表拆分为原链表和复制后的链表
currentNode = pHead;
RandomListNode pCloneHead = pHead.next;
while(currentNode != null) {
RandomListNode cloneNode = currentNode.next;
currentNode.next = cloneNode.next;
cloneNode.next = cloneNode.next==null?null:cloneNode.next.next;
currentNode = currentNode.next;
}
return pCloneHead;
}
}