- 輸入一個複雜鏈表(每個節點中有節點值,以及兩個指針,一個指向下一個節點,另一個特殊指針指向任意一個節點),返回結果爲複製後複雜鏈表的head。(注意,輸出結果中請不要返回參數中的節點引用,否則判題程序會直接返回空)
- 我的算法思想很簡單,先複製next指針,然後採用類似冒泡算法複製random指針。
/*
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)
return NULL;
RandomListNode* ans = (RandomListNode*)malloc(sizeof(RandomListNode));
ans->label = pHead->label;
ans->next=ans->random = NULL;
if(!pHead->next)
return ans;
RandomListNode* p =pHead->next;
RandomListNode* p_ans = ans;
while(p)
{
RandomListNode* node = (RandomListNode*)malloc(sizeof(RandomListNode));
node->label=p->label;
node->random=NULL;
node->next=NULL;
p_ans->next=node;
p_ans = node;
p=p->next;
}
p=pHead;
p_ans = ans;
while(p)
{
RandomListNode* p1 = pHead;
RandomListNode* p2 = ans;
while(p->random != p1)
{
p1=p1->next;
p2=p2->next;
}
p_ans->random=p2;
p=p->next;
p_ans=p_ans->next;
}
return ans;
}
};