当前位置: 首页 > 文档资料 > 算法珠玑 >

linear-list/linked-list/copy-list-with-random-pointer

优质
小牛编辑
120浏览
2023-12-01

Copy List with Random Pointer

描述

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

分析

代码

// Copy List with Random Pointer
// 两遍扫描,时间复杂度O(n),空间复杂度O(1)
class Solution {
    public RandomListNode copyRandomList(RandomListNode head) {
        for (RandomListNode cur = head; cur != null; ) {
            RandomListNode node = new RandomListNode(cur.label);
            node.next = cur.next;
            cur.next = node;
            cur = node.next;
        }

        for (RandomListNode cur = head; cur != null; ) {
            if (cur.random != null)
                cur.next.random = cur.random.next;
            cur = cur.next.next;
        }

        // 分拆两个单链表
        RandomListNode dummy = new RandomListNode(-1);
        for (RandomListNode cur = head, new_cur = dummy;
                cur != null; ) {
            new_cur.next = cur.next;
            new_cur = new_cur.next;
            cur.next = cur.next.next;
            cur = cur.next;
        }
        return dummy.next;
    }
};
// Copy List with Random Pointer
// 两遍扫描,时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
    RandomListNode *copyRandomList(RandomListNode *head) {
        for (RandomListNode* cur = head; cur != nullptr; ) {
            RandomListNode* node = new RandomListNode(cur->label);
            node->next = cur->next;
            cur->next = node;
            cur = node->next;
        }

        for (RandomListNode* cur = head; cur != nullptr; ) {
            if (cur->random != NULL)
                cur->next->random = cur->random->next;
            cur = cur->next->next;
        }

        // 分拆两个单链表
        RandomListNode dummy(-1);
        for (RandomListNode* cur = head, *new_cur = &dummy;
                cur != nullptr; ) {
            new_cur->next = cur->next;
            new_cur = new_cur->next;
            cur->next = cur->next->next;
            cur = cur->next;
        }
        return dummy.next;
    }
};