leetcode -- Copy List with Random Pointer

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.

[解题思路]

1.和clone graph那题差不多，使用一个map来保存已经创建好的node。

第一步：先复制原始链表中的节点，用next指针链接起来，在复制节点时放入map中

第二步：设置每个节点的random指针

空间复杂度为O(n)，时间复杂度为O(n)

Accepted Code

public RandomListNode copyRandomList(RandomListNode head) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == null){
            return head;
        }
        
        Map<RandomListNode, RandomListNode> map = new HashMap<RandomListNode, RandomListNode>();
        RandomListNode newHead = new RandomListNode(head.label);
        map.put(head, newHead);
        RandomListNode p1 = head;
        RandomListNode p2 = newHead;
        p1 = p1.next;
        // p2 = p2.next;
        
        while(p1 != null){
            RandomListNode tmp = new RandomListNode(p1.label);
            p2.next = tmp;
            map.put(p1, tmp);
            p1 = p1.next;
            p2 = p2.next;
        }        
        
        p1 = head;
        p2 = newHead;
        while(p1 != null){
            RandomListNode random = p1.random;
            if(random != null)
                p2.random = map.get(random);
            p1 = p1.next;
            p2 = p2.next;
        }
       
        return newHead;
    }

2.上面的做法使用map的意义在于设置random指针时可以直接从map中找到，但使用map会带来O(n)的空间消耗

换个思路：第一步仍然根据原始链表的每个结点n创建相应的n'。将n'链接到n后面

第二步设置n'的random指针。n'的random指针所指向的结点是n的random指针所指向的结点的下一个结点

第三步把第二步得到的链表拆成两个链表

public class Solution {
    
    public RandomListNode cloneNodes(RandomListNode head){
        RandomListNode p1 = head;
        while(p1 != null){
            RandomListNode c1 = new RandomListNode(p1.label);
            c1.next = p1.next;
            p1.next = c1;
            p1 = c1.next;
        }
        return head;
    }
    
    public RandomListNode setRandomPointer(RandomListNode head){
        RandomListNode p1 = head;
        while(p1 != null){
            RandomListNode r1 = p1.random;
            if(r1 != null){
                p1.next.random = r1.next;
            }
            p1 = p1.next.next;
        }
        return head;
    }
    
    public RandomListNode reconnectNodes(RandomListNode head){
        RandomListNode cloneHead = head.next;
        RandomListNode p1 = head;
        RandomListNode p2 = cloneHead;
        while(p1 != null && p2 != null){
            p1.next = p2.next;
            p1 = p1.next;
            if(p1 != null){
34                 p2.next = p1.next;
35             }
            p2 = p2.next;
        }
        return cloneHead;
    }
    
    public RandomListNode copyRandomList(RandomListNode head) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == null){
            return head;
        }
        
        cloneNodes(head);
        setRandomPointer(head);
        return reconnectNodes(head);
    }
}

ATT：拆成两个链表时，当处理到最后一个节点时需要注意