LeetCode 138. Copy List with Random Pointer

原题链接在这里：https://leetcode.com/problems/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.

题解：

每个node 带了一个random指针，问题就在于copy时也许这个random指针指向的还没有遍历到。

第一种方法使用HashMap 做两遍itertion, 第一遍复制原有list包括next, 但是不复制random, 把original node 和 copy node 成对放到HashMap中，original node是键。第二遍遍历时把copy的random指向oringal node random 在HashMap中对应的点。

如此Time Complexity: O(n). Space O(n), 这种方法就不写了。

第二种方法不利用额外空间，那么只能利用原有数据结构。一共需要做三次iteration, 第一次遍历时对于每一个original node 做 copy 然后insert 到original node的next里。第二次遍历时如果origianl的random不指向null, 就让它后面的copy node random 指向原有random指向点的next, cur.next.random = cur.random.next, 因为random指向点的next 就是指向点的copy. 第三次遍历时把原有list 拆成两个list, 一个原有list, 一个复制list.

Note: 做第二次遍历时要注意original random指向null的情况，因为后面用到了cur.random.next 若是指向null, 是没有next的.

Time Complexity: O(n). Space O(1).

AC Java:

/**
 * 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 {
    public RandomListNode copyRandomList(RandomListNode head) {
        if(head == null){
            return null;
        }
        
        //First iteration, copy node and append it to the original node
        RandomListNode cur = head;
        while(cur != null){
            RandomListNode copy = new RandomListNode(cur.label);
            copy.next = cur.next;
            cur.next = copy;
            cur = cur.next.next;
        }
        
        //Second iteration, if original random doesn't point to null, assign corresponding copy node random  point to origianl 
        //random's next node
        cur = head;
        while(cur != null){
            if(cur.random != null){
                cur.next.random = cur.random.next;
            }
            
            cur = cur.next.next;
        }
        
        //Third iteration, separate list into two. One is origianl list, the other is the copy list.
        cur = head;
        RandomListNode copyHead = cur.next;
        RandomListNode copyCur = copyHead;
        while(cur != null) {
            cur.next = cur.next.next;
            copyCur.next = copyCur.next != null ? copyCur.next.next : null;
            
            cur = cur.next;
            copyCur = copyCur.next;
        }
        
        return copyHead;
    }
}

DFS 可以像 Clone Graph.

Time Complexity: O(N+E). Space: O(N).

public class Solution {
    Map<Integer, RandomListNode> visited = new HashMap<Integer, RandomListNode>();
    
    public RandomListNode copyRandomList(RandomListNode head) {
        if(head == null){
            return head;
        }
        if(visited.containsKey(head.label)){
            return visited.get(head.label);
        }
        
        RandomListNode copy = new RandomListNode(head.label);
        visited.put(head.label, copy);
        
        copy.next = copyRandomList(head.next);
        copy.random = copyRandomList(head.random);
        return copy;
    }
}