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

題目

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. 跟複製常規的單向鏈表的區別主要在於每個結點多了個random指針。
2. 可以先生成一個random指針均爲NULL的新鏈表，並保存鏈表對應位置新舊鏈表指針的一一映射關係(map)，則第二次遍歷新鏈表時可以爲所有的random指針賦值。

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

參考：http://blog.csdn.net/litoupu/article/details/41623807

技巧：直接在原有鏈表上進行操作，每個結點後都new一個新的同值結點並連接。第二次再將random值通過random->next賦值。最後分離出原有鏈表和新鏈表。