单链表OJ题——11.随机链表的复制

/*
解题思路：
此题可以分三步进行：
1.拷贝链表的每一个节点，拷贝的节点先链接到被拷贝节点的后面
2.复制随机指针的链接：拷贝节点的随机指针指向被拷贝节点随机指针的下一个位置
3.拆解链表，把拷贝的链表从原链表中拆解出来
*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        // 1.拷贝链表，并插入到原节点的后面
        Node* cur = head;
        while(cur)
        {
            Node* next = cur->next;

            Node* copy = (Node*)malloc(sizeof(Node));
            copy->val = cur->val;

            // 插入
            cur->next = copy;
            copy->next = next;

            // 迭代往下走
            cur = next;
        }

        // 2.置拷贝节点的random
        cur = head;
        while(cur)
        {
            Node* copy = cur->next;
            if(cur->random != NULL)
                copy->random = cur->random->next;
            else
                copy->random = NULL;

            cur = copy->next;
        }

        // 3.解拷贝节点，链接拷贝节点
        Node* copyHead = NULL, *copyTail = NULL;
        cur = head;
        while(cur)
        {
            Node* copy = cur->next;
            Node* next = copy->next;

            // copy解下来尾插
            if(copyTail == NULL)
            {
                copyHead = copyTail = copy;
            }
            else
            {   
                copyTail->next = copy;
                copyTail = copy;
            }

            cur->next = next;

            cur = next;
        }

        return copyHead;
    }
};