复杂链表的复制

Desicription

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)

Solution

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/*
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 == nullptr) {
return nullptr;
}
for(RandomListNode* now = pHead; now != nullptr; now = now->next) {
auto newNode = new RandomListNode(now->label);
newNode->next = now->random;
now->random = newNode;
}
for(RandomListNode* now = pHead; now != nullptr; now = now->next) {
auto newNode = now->random;
newNode->random = newNode->next != nullptr ? newNode->next->random : nullptr;
}
auto newHead = pHead->random;
for(RandomListNode* now = pHead; now != nullptr; now = now->next) {
auto newNode = now->random;
now->random = newNode->next;
newNode->next = now->next != nullptr ? now->next->random : nullptr;
}
return newHead;
}
};