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.

Could you solve it with O(1) space?

1. O(n)空間：和deep copy graph類似，一開始建一個Map，映射原始點和新創點，之後遍歷過去一個個把原始的邊在新的鏈表裏連上。a.造節點並在map裏輸入舊新結點的映射關系 b.根據舊結點信息在新結點中把鏈條關系連上

2.O(1)空間：巧妙三步走：a.創造新節點緊跟在屁股後。1-1‘-2-2‘-3-3‘-4-4‘-null。b.把新的random指針連好(crt.next.random = crt.random.next) c.把兩個纏在一起的解旋

細節：1.小心crt.random是null的情況，小心head是null的情況，小心最後4-4‘-null時crt指到null就不能延續後指crt.next的情況，總之這題裏很多null指針報錯的地方，你一定把所有用到xx.null的地方都檢查一遍，看xx會不會是null。2.解旋前先把新鏈表的頭地址提出來，避免解旋後你用head.next已經再也不能索引到它的情況。

法1.O(n)空間

/**
 * Definition for singly-linked list with a random pointer.
 * class RandomListNode {
 *     int label;
 *     RandomListNode next, random;
 *     RandomListNode(int x) { this.label = x; }
 * };
 */
public class Solution {
    /**
     * @param head: The head of linked list with a random pointer.
     * 
 
@return: A new head of a deep copy of the list.
     */
    public RandomListNode copyRandomList(RandomListNode head) {
        // write your code here
        Map<RandomListNode, RandomListNode> map = new HashMap<>();
        
        // copy the node and make connection in map
        RandomListNode crt = head;
        map.put(null, null);
        while (crt != null) {
            RandomListNode copy = new RandomListNode(crt.label);
            map.put(crt, copy);
            crt = crt.next;
        }
        
        // use map to link the random and next in the new linkedList
        crt = head;
        while (crt != null) {
            RandomListNode crtCopy = map.get(crt);
            crtCopy.next = map.get(crt.next);
            crtCopy.random = map.get(crt.random);
            crt = crt.next;
        }
        
        return map.get(head);
    }
}

法2.O(1)空間

/**
 * Definition for singly-linked list with a random pointer.
 * class RandomListNode {
 *     int label;
 *     RandomListNode next, random;
 *     RandomListNode(int x) { this.label = x; }
 * };
 */
public class Solution {
    /**
     * @param head: The head of linked list with a random pointer.
     * @return: A new head of a deep copy of the list.
     */
    public RandomListNode copyRandomList(RandomListNode head) {
        // write your code here
        
        RandomListNode crt = head;
        while (crt != null) {
            RandomListNode temp = crt.next;
            crt.next = new RandomListNode(crt.label);
            crt.next.next = temp;
            crt = temp;
        }
        
        crt = head;
        while (crt != null) {
            // 小心crt.random是null的情況！
            crt.next.random = crt.random == null ? null : crt.random.next;
            crt = crt.next.next;
        }
        
        crt = head;
        // 小心head是null的情況，用到xx.next的地方都註意一下xx會不會是null出錯
        RandomListNode copy = crt == null ? crt : crt.next;
        // 要提早把這個頭存出來以便後續返回
        RandomListNode copyHead = head.next;
        while (crt != null) {
            RandomListNode temp = crt.next.next;
            crt.next = temp;
            // 拆開的時候小心null情況，因為你兩條路要從一個null拆兩個null出來
            // 這裏不處理會報空指針，不能讀取null.next
            copy.next = temp == null ? temp : temp.next;
            crt = temp;
            copy = crt == null ? crt : crt.next;
        }
        
        return copyHead;
    }
}

lintcode105- Copy List with Random Pointer- medium