題目描述
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)
題目分析
在牛客網討論區的圖:
這樣的好處是在第二遍遍歷時,複製random比較方便,寫爲
node->random = p->random->next;
/*
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 nullptr;
RandomListNode *p, *n, *nHead, *node;
p = pHead;
while(p){
node = new RandomListNode(p->label);
node->next = p->next;
p->next = node;
p = node->next;
}
p = pHead;
//while(n){
//n->random = p->random->next;
//p = n->next;
//n = p->next;
//出錯:A->A',則p = null, n = null->next導致錯誤
//}
while(p){
n = p->next;
if(p->random) //保證random不是指向空節點
n->random = p->random->next;
p = n->next;
}
p = pHead;
n = p->next;
nHead = n;
//此處條件爲p->next,當A->A'時,若條件爲n,將無法拆分
while(p->next){
n = p->next;
p->next = n->next;
//p = p->next;
//n->next = p->next;
//n = n->next;
p = n;
}
return nHead;
}
};