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

题目

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.

分析

1. 跟复制常规的单向链表的区别主要在于每个结点多了个random指针。
2. 可以先生成一个random指针均为NULL的新链表，并保存链表对应位置新旧链表指针的一一映射关系(map)，则第二次遍历新链表时可以为所有的random指针赋值。

Coding

/**
 * 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) {
        if (!head)
            return NULL;
        // new list head
        RandomListNode* newHead = new RandomListNode(head->label);
        map<RandomListNode*, RandomListNode*> MapListNode;
        MapListNode[head] = newHead;
        // new list node
        RandomListNode* oldNode = head->next;
        RandomListNode* tempNode = newHead;
        // generate new list
        while (oldNode) {
            RandomListNode* newNode = new RandomListNode(oldNode->label);
            tempNode->next = newNode;
            MapListNode[oldNode] = newNode;
            oldNode = oldNode->next;
            tempNode = newNode;
        }
        // assign random pointer value
        tempNode = newHead;
        oldNode = head;
        while (oldNode && tempNode) {
            if (oldNode->random) {
                tempNode->random = MapListNode[oldNode->random];
            }
            oldNode = oldNode->next;
            tempNode = tempNode->next;
        }
        return newHead;
    }
};

拓展

参考：http://blog.youkuaiyun.com/litoupu/article/details/41623807

技巧：直接在原有链表上进行操作，每个结点后都new一个新的同值结点并连接。第二次再将random值通过random->next赋值。最后分离出原有链表和新链表。