剑指offer 35

来源:剑指 Offer 35. 复杂链表的复制

/*
// 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;
        /* 1. 复制各节点，并构建拼接链表
        *  插入节点常规操作，注意cur = tmp.next;
        */
        while(cur != null) {
            Node tmp = new Node(cur.val);
            tmp.next = cur.next;
            cur.next = tmp;
            cur = tmp.next;
        }
        
        /* 2. 构建各新节点的 random 指向
        *   复制节点的random与原节点指向同一个
        */
        cur = head;
        while(cur != null) {
            if(cur.random != null)
                cur.next.random = cur.random.next;
            cur = cur.next.next;
        }
        /* 3. 拆分两链表
        *   cur是用来找复制点的
        *   pre是用来找原节点的
        *   res指向复制点的头，用来输出复制点的 
        */
        cur = head.next;
        Node pre = head, res = 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; // 单独处理原链表尾节点
        return res;      // 返回新链表头节点
    }
}