138. 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.

分析：

这题的关键其实是如何copy random pointer，找到一个random pointer指向的node, time complexity is O(n). 所以总的复杂度在O(n^2).

这里有一个比较巧妙的方法。在每个Node后面添加一个Node，新node的值和前一个Node一样。这样的话random pointer就很容易copy了，

利用hashmap保存original和copy

/*
// Definition for a Node.
class Node {
    int val;
    Node next;
    Node random;

    public Node(int val) {
        this.val = val;
        this.next = null;
        this.random = null;
    }
}
*/

class Solution {
    public Node copyRandomList(Node head) {
        if (head == null) return null;
        Map<Node, Node> map = new HashMap<>();
        Node current = head;
        while(current != null) {
            map.put(current, new Node(current.val));
            current = current.next;
        }         
        Node firstNode = head;
        while(head != null) {
            Node temp = map.get(head);
            temp.next = map.get(head.next);
            if(head.random != null) {
                temp.random = map.get(head.random);
            }
            head = head.next;
        }
        return map.get(firstNode);
    }
}

 

 

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