138. Copy List with Random Pointer

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Problem:

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.

The Linked List is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

• val: an integer representing Node.val
• random_index: the index of the node (range from 0to n-1) where random pointer points to, or null if it does not point to any node.

Example 1:

Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2:

Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]]

Example 3:

Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]

Example 4:

Input: head = []
Output: []
Explanation: Given linked list is empty (null pointer), so return null.

Constraints:

• -10000 <= Node.val <= 10000
• Node.random is null or pointing to a node in the linked list.
• Number of Nodes will not exceed 1000.

思路：

Solution:

Node* copyRandomList(Node* head) {
    Node *newHead, *l1, *l2;
    if (head == NULL)  return NULL;
    for (l1 = head; l1 != NULL; l1 = l1->next->next) {
        l2 = new Node(l1->val);
        l2->next = l1->next;
        l1 -> next = l2;
    }
    
    newHead = head->next;
    for (l1 = head; l1 != NULL; l1 = l1->next->next) {
        if (l1->random != NULL)  l1->next->random = l1->random->next;
    }
    for (l1 = head; l1 != NULL; l1 = l1->next) {
        l2 = l1->next;
        l1->next = l2->next;
        if (l2->next != NULL)  l2->next = l2->next->next;
    }
    return newHead;
}

性能:
Runtime: 12 ms??Memory Usage: 13.3 MB

138. Copy List with Random Pointer

标签：present   ted   dia   lis   dom   repr   ber   solution   tps   

原文地址：https://www.cnblogs.com/dysjtu1995/p/12275040.html