剑指Offer_编程题_25

题目描述

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。(注意,输出结果中请不要返回参数中的节点引用,否则判题程序会直接返回空)
/*
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 == NULL){
            return NULL;
        }
        map<RandomListNode*,RandomListNode*>tmpMap;
        RandomListNode* p1 = NULL;
        RandomListNode* p2Head = NULL;
        RandomListNode* tmpHead = pHead;
        if(tmpHead){
            p1 = new RandomListNode(tmpHead->label);
            p2Head = p1;
            tmpMap[tmpHead] = p1;
            tmpHead = tmpHead->next;
        }
        while(tmpHead){
            RandomListNode* tmp = new RandomListNode(tmpHead->label);
            p1->next = tmp;
            p1 = tmp;
            tmpMap[tmpHead] = tmp;
            tmpHead = tmpHead->next;
        }
        tmpHead = pHead;
        map<RandomListNode*,RandomListNode*>::iterator it;
        map<RandomListNode*,RandomListNode*>::iterator it_random;
        while(tmpHead){
            if(tmpHead->random){
                it = tmpMap.find(tmpHead);
                if(it != tmpMap.end()){
                    it_random = tmpMap.find(tmpHead->random);
                    if(it_random != tmpMap.end()){
                        it->second->random = it_random->second;
                    }
                }
                
            }
            tmpHead = tmpHead->next;
        }
        return p2Head;
    }
};

  

posted @ 2018-05-05 15:39  gaoren  阅读(162)  评论(0编辑  收藏  举报