【剑指offer】复杂链表的复制（链表）

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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）

链接

https://www.nowcoder.com/practice/f836b2c43afc4b35ad6adc41ec941dba?tpId=13&tqId=11178&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

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* head = pHead;
        while(head){
        	RandomListNode* temp = (RandomListNode*)malloc(sizeof(RandomListNode));
        	temp->label = head->label;
        	RandomListNode* next = head->next;
        	head->next = temp;
        	temp->next = next;
        	head = next;
        }
        head = pHead;
        while(head){
        	if(head->next){
        		if(head->random){
        			head->next->random = head->random->next;
        		}
        		else{
        			head->next->random = head->random;
        		}
        		head = head->next->next;
        	}
        }
        // 奇数节点串起来是原来的，偶数节点串起来是需要的
//        RandomListNode* ans = pHead->next;
//        head = ans;
//		while(head){
//			if(head->next){
//				head->next = head->next->next;
//			}
//			head = head->next;
//		}
		// 原链表需要还原，否则错误
		head = pHead;
		RandomListNode* ans = head->next;
		RandomListNode* pClone = ans;
		while(head){
			if(head->next){
				head->next = head->next->next;
			}
			if(pClone->next){
				pClone->next = pClone->next->next;
			}
			head = head->next;
			pClone = pClone->next;
		}
		
		return ans;
    }
};

/*
int main()
{
	RandomListNode *root = new RandomListNode(1);
	RandomListNode *r1 = new RandomListNode(2);
	RandomListNode *r2 = new RandomListNode(3);
	RandomListNode *r3 = new RandomListNode(4);
	RandomListNode *r4 = new RandomListNode(5);
	root->next = r1;
	r1->next = r2;
	r2->next = r3;
	r3->next = r4;
	
	root->random = r2;
	r1->random = r4;
	r2->random = NULL;
	r3->random = r1;
	r4->random = NULL;
	
	Solution *s = new Solution();
	RandomListNode* ans, *head;
	ans = s->Clone(root);
    head = ans;
    while(head){
    	cout << head->label << " ";
		head = head->next; 
    }
	return 0;
}
*/