题解 | #复杂链表的复制#

/*
struct RandomListNode {
    int label;
    struct RandomListNode *next, *random;
    RandomListNode(int x) :
            label(x), next(NULL), random(NULL) {
    }
};
*/
class Solution {
  public:
    RandomListNode* Clone(RandomListNode* pHead) {
        // 空的判断
        if (pHead == nullptr)
            return nullptr;

        int listLen;
        RandomListNode* pReHead = new RandomListNode(pHead->label);
        RandomListNode* pInCurr = pHead;
        RandomListNode* pReCurr = pReHead;

        // 先复制主干next 上的
        pInCurr = pInCurr->next;
        for (listLen = 0; pInCurr != nullptr; listLen++) {
            pReCurr->next = new RandomListNode(pInCurr->label);
            pInCurr = pInCurr->next;
            pReCurr = pReCurr->next;
        }

        // 再复制跳跃的
        int value;
        pInCurr = pHead;
        pReCurr = pReHead;
        RandomListNode* pReScan = pReHead;

        for (int i = 0; i < listLen ; i++ ) {
            if (pInCurr->random != nullptr) {
                // 将输出的扫描指针归位
                pReScan = pReHead;
                // 提取出对应节点的值
                value = pInCurr->random->label;
                for (; pReScan != nullptr;) {
                    if (pReScan->label == value) {
                        pReCurr->random = pReScan;
                    }
                    pReScan = pReScan->next;
                }

            }
            pInCurr = pInCurr->next;
            pReCurr = pReCurr->next;
        }
        return pReHead;
    }
};