剑指 Offer 35. 复杂链表的复制

1. Question

请实现 copyRandomList 函数,复制一个复杂链表。在复杂链表中,每个节点除了有一个 next 指针指向下一个节点,还有一个 random 指针指向链表中的任意节点或者 null

2. Examples

示例 1:

img
Image
输入:head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
输出:[[7,null],[13,0],[11,4],[10,2],[1,0]]

示例 2:

img
Image
输入:head = [[1,1],[2,1]]
输出:[[1,1],[2,1]]

示例 3:

img
Image
输入:head = [[3,null],[3,0],[3,null]]
输出:[[3,null],[3,0],[3,null]]

示例 4:

输入:head = []
输出:[]
解释:给定的链表为空(空指针),因此返回 null。

3. Constraints

  • -10000 <= Node.val <= 10000
  • Node.random 为空(null)或指向链表中的节点。
  • 节点数目不超过 1000 。

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/fu-za-lian-biao-de-fu-zhi-lcof 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

通过HashMap映射新旧的Node。

/*
// 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 oldHead = head;
    Node newHead = new Node(oldHead.val);
    Node ans = newHead;

    HashMap<Node, Node> map = new HashMap<>();
    map.put(oldHead, newHead);

    while(oldHead.next != null) {
      Node node = new Node(oldHead.next.val);
      newHead.next = node;
      oldHead = oldHead.next;
      newHead = newHead.next;
      map.put(oldHead, newHead);
    }

    newHead = ans;
    oldHead = head;

    while(oldHead != null) {
      newHead.random = map.get(oldHead.random);
      newHead = newHead.next;
      oldHead = oldHead.next;
    }


    return ans;
  }
}
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