LeetCode || Copy List with Random Pointer
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
Return a deep copy of the list.
思路1:最傻瓜的方法是首先遍歷一次建立next關系的新list。然後第二次遍歷處理random關系,對於每個有random結點的node,我們都從表頭開始遍歷尋找其random的結點。然後給新list的相應結點賦值。這種話對於每個node,尋找random須要花費O(N)時間。故總時間復雜度為O(N^2)。這個方案會超時。
思路2:改進思路1。假設處理random關系的復制,使其復雜度降為O(N)?答案是要找到原node的random指向的結點在新list中相應的那個結點,假設能一下找到,那麽就攻克了;實現方法是使用一個map<old, new>。記錄原node與新node的相應關系。然後進行兩次遍歷,第一次建立next關系的新list。第二次給新list建立random指向關系;代碼例如以下:
/** * Definition for singly-linked list with a random pointer. * struct RandomListNode { * int label; * RandomListNode *next, *random; * RandomListNode(int x) : label(x), next(NULL), random(NULL) {} * }; */ class Solution { public: RandomListNode *copyRandomList(RandomListNode *head) { map<RandomListNode *, RandomListNode *> mapNodes; RandomListNode *newList = NULL; RandomListNode *newHead = NULL; RandomListNode *p = head; if(head==NULL) return NULL; while(p!=NULL){ RandomListNode *q = new RandomListNode(p->label); mapNodes[p] = q; if(newHead==NULL){ newHead = q; newList = q; } else{ newList->next = q; newList = newList->next; } p = p->next; } p = head; newList = newHead; while(p!=NULL){ if(p->random!=NULL) newList->random = mapNodes[p->random]; //註意這裏要用p->random而不是p p = p->next; newList = newList->next; } return newHead; //返回值是newHead而不是newList,由於此時newList指向尾部 } };
思路3:思路2沒有改變原list結構,可是使用了map,故須要額外的內存空間,假設原list解構可變。那麽能夠不必使用map記錄映射關系,而是直接把復制的node放在原node的後面,這樣結構變為:
上面為第一次遍歷,第二次遍歷時把紅色的新node的random域賦值。規則是:
newNode->ranodm = oldNode->random->next;
然後第三次遍歷把上面的鏈表拆分為兩個就可以。代碼例如以下:
/** * Definition for singly-linked list with a random pointer. * struct RandomListNode { * int label; * RandomListNode *next, *random; * RandomListNode(int x) : label(x), next(NULL), random(NULL) {} * }; */ class Solution { public: RandomListNode *copyRandomList(RandomListNode *head) { RandomListNode *newList = NULL; RandomListNode *newHead = NULL; RandomListNode *p = head; if(head==NULL) return NULL; while(p!=NULL){ RandomListNode *q = new RandomListNode(p->label); q->next = p->next; p->next = q; p = q->next; } p = head; while(p!=NULL){ if(p->random != NULL) p->next->random = p->random->next; p = p->next->next; } p = head; while(p!=NULL && p->next!=NULL){ //註意這裏的條件,首先要推斷p是否為null if(p==head){ newHead = p->next; newList = newHead; } else{ newList->next = p->next; newList = newList->next; } p->next = newList->next; p = p->next; } return newHead; //返回值是newHead而不是newList,由於此時newList指向尾部 } };
參考鏈接:http://www.2cto.com/kf/201310/253477.html
LeetCode || Copy List with Random Pointer