Offer.P35[复杂链表的复制]

方法一：哈希表

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/*
// 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;
        Node p = head;
        Map<Node, Node> map = new HashMap<>();
        while (p != null) {
            map.put(p, new Node(p.val));
            p = p.next;
        }
        p = head;
        while (p != null) {
            map.get(p).next = map.get(p.next);
            map.get(p).random = map.get(p.random);
            p = p.next;
        }
        return map.get(head);
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：拼接+拆分

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/*
// 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;
        Node cur = head;
        // 复制各节点，并构建新链表
        while (cur != null) {
            Node temp = new Node(cur.val);
            temp.next = cur.next;
            cur.next = temp;
            cur = temp.next;
        }
        cur = head;
        // 构建新链表的random指向
        while (cur != null) {
            if (cur.random != null) {
                cur.next.random = cur.random.next;
            }
            cur = cur.next.next;
        }
        // 拆分链表
        cur = head.next;
        Node pre = head, ans = head.next;
        while (cur.next != null) {
            pre.next = pre.next.next;
            cur.next = cur.next.next;
            pre = pre.next;
            cur = cur.next;
        }
        pre.next = null;
        return ans;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(1)$