剑指 Offer 35. 复杂链表的复制

剑指 Offer 35. 复杂链表的复制

方法一

哈希表+两次遍历

class Solution {
    public Node copyRandomList(Node head) {
        Node dummy = new Node(-1), h = head, d = dummy;
        HashMap<Node, Node> map = new HashMap<>();
        
        while(h != null){
            d.next = new Node(h.val);
            map.put(h, d.next);
            h = h.next;
            d = d.next;
        }

        h = head;
        d = dummy.next;
        while(h != null){
            d.random = map.get(h.random);
            h = h.next;
            d = d.next;
        }

        return dummy.next;
    }
}

方法二

拼接与拆分+三次遍历 不使用额外空间

class Solution {
    public Node copyRandomList(Node head) {
        if(head == null) return null;
        Node dummy = new Node(-1), h = head, d = dummy;

        // 在每个节点后面复制一次
        while(h != null){
            Node tmp = new Node(h.val);
            tmp.next = h.next;
            h.next = tmp;
            h = h.next.next;
        }

        // 修改复制节点的random指针
        h = head;
        while(h != null){
            if(h.random != null) h.next.random = h.random.next;
            h = h.next.next;
        }

        // 拆分
        h = head;
        dummy.next = h.next;
        d = h.next;
        while(h.next.next != null){  // 最后一个节点不处理了
            h.next = h.next.next;
            d.next = d.next.next;
            h = h.next;
            d = d.next;
        }
        h.next = null;  // 单独处理最后一个节点

        return dummy.next;
    }
}

方法三

回溯 + 哈希表

class Solution {
    HashMap<Node, Node> map = new HashMap<>();

    public Node copyRandomList(Node head) {
        if(head == null) return null;
        if(!map.containsKey(head)){
            Node newHead = new Node(head.val);
            map.put(head, newHead);
            newHead.next = copyRandomList(head.next);
            newHead.random = copyRandomList(head.random);
        }
        return map.get(head);
    }
}