题目
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。
思路一
我们可以用哈希表来存储<N,N'>,然后连接next和random
用空间复杂度O(n)将原本时间复杂度为O(n2)降低到O(n)
class Solution { public: RandomListNode* Clone(RandomListNode* pHead) { // 哈希表法 if (!pHead) return nullptr; unordered_map<RandomListNode*, RandomListNode*> table; for (RandomListNode* p = pHead; p != nullptr; p = p->next) table[p] = new RandomListNode(p->label); for (RandomListNode* p = pHead; p != nullptr; p = p->next) { table[p]->next = table[p->next]; table[p]->random = table[p->random]; } return table[pHead]; } };
思路二
三步走
1、将链表变为 A->A'->B->B'->C->C'
2、复制random连接
3、将这个链表按奇偶顺序拆成两个链表
/* 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* curNode = pHead; // 复制链表形成 A - A' - B - B' - C - C'- NULL while (curNode) { RandomListNode* pClone = new RandomListNode(curNode->label); pClone->next = curNode->next; curNode->next = pClone; curNode = pClone->next; } curNode = pHead; //复制random while (curNode) { RandomListNode* pClone = curNode->next; if (curNode->random) pClone->random = curNode->random->next; curNode = pClone->next; } // 拆分链表 curNode = pHead; RandomListNode* pCloneHead = pHead->next, *temp = nullptr; while (curNode->next) { temp = curNode->next; curNode->next = temp->next; curNode = temp; } return pCloneHead; } };