[leetcode]copy-list-with-random-pointer

思路

链表里面的顺序是个很麻烦的事，如果要复制的时候保留相互关系，以原链表为基础，当做一个模子，
直接在上面弄一份一份副本，就像做一把钥匙，倒个模子。又或者像草履虫，无性繁殖，biaji，生出一个新的。

Solution 1

/**
 * 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) {
        head = insertAfterNode(head);
        head = copyRandomity(head);
        RandomListNode* head2 = asexualPropagation(head);
        return head2;
    }


private:
    //在所有的node之后插入一个label相同的node
    RandomListNode* insertAfterNode(RandomListNode *head){
        RandomListNode* tempHead = head;
        while(head){
            RandomListNode *node = new RandomListNode(head->label);

            //insert node after head , label it 
            RandomListNode* temp = head->next;
            head->next = node;
            node->next = temp;

            head = temp;
        }

        return tempHead;
    }

    //复制随机关系
    RandomListNode* copyRandomity(RandomListNode* head){
        RandomListNode* tempHead = head;
        while(head){
            //head's random node
            if(head->random){
                RandomListNode* randomNode = head->random;
                head->next->random = randomNode->next;
            }

            head = head->next->next; //skip one every time， head->next must not be null
        }
        return tempHead;
    }

    //分离
    RandomListNode* asexualPropagation(RandomListNode* head){
        //check
        if(!head){
            return head;
        }

        RandomListNode* head2 = head->next;    
        while(head->next->next){ //最后到达整个的倒数第二个元素
            RandomListNode* nextOldNode = head->next->next;

            //link the new list 
            RandomListNode* thisNewNode = head->next;
            RandomListNode* nextNewNode = head->next->next->next;
            thisNewNode->next = nextNewNode;

            //link the old list
            head->next = nextOldNode;


            head = nextOldNode; //skip one every time， head->next must not be null
        }
        head->next = nullptr; //最后一个oldNode要结尾， 而newNode不用，因为next已经是null

        return head2;

    }
};