复杂链表的复制

题解1: 哈希表

空间和时间都是O(n)

class Solution {
    public Node copyRandomList(Node head) {
        if (head == null) {
            return head;
        }
        //map中存的是(原节点，拷贝节点)的一个映射
        Map<Node, Node> map = new HashMap<>();
        for (Node cur = head; cur != null; cur = cur.next) {
            map.put(cur, new Node(cur.val));
        }
        //将拷贝的新的节点组织成一个链表
        for (Node cur = head; cur != null; cur = cur.next) {
            map.get(cur).next = map.get(cur.next);
            map.get(cur).random = map.get(cur.random);
        }

        return map.get(head);
    }
}

题解2: 原地修改

空间O(1)

class Solution {
    public Node copyRandomList(Node head) {
        if (head == null) {
            return head;
        }
        //将拷贝节点放到原节点后面，例如1->2->3这样的链表就变成了这样1->1'->2'->3->3'
        for (Node node = head, copy = null; node != null; node = node.next.next) {
            copy = new Node(node.val);
            copy.next = node.next;
            node.next = copy;
        }
        //把拷贝节点的random指针安排上
        for (Node node = head; node != null; node = node.next.next) {
            if (node.random != null) {
                node.next.random = node.random.next;
            }
        }
        //分离拷贝节点和原节点，变成1->2->3和1'->2'->3'两个链表，后者就是答案
        Node newHead = head.next;
        for (Node node = head, temp = null; node != null && node.next != null;) {
            temp = node.next;
            node.next = temp.next;
            node = temp;
        }

        return newHead;
    }
}

题解3: DFS

图的基本单元是 顶点，顶点之间的关联关系称为 边，我们可以将此链表看成一个图

class Solution:
    def copyRandomList(self, head: 'Node') -> 'Node':
        def dfs(head):
            if not head: return None
            if head in visited:
                return visited[head]
            # 创建新结点
            copy = Node(head.val, None, None)
            visited[head] = copy
            copy.next = dfs(head.next)
            copy.random = dfs(head.random)
            return copy
        visited = {}
        return dfs(head)