138-复制带随机指针的链表

LeetCode 138. Copy List with Random Pointer

LeetCode-138

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

Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

Return the head of the copied linked 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 0 to n-1) that the random pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.

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]]

method: 哈希表

第一次遍历只生成节点，用哈希表记录原节点和新节点的对应关系

然后就可以用hash[cur->next]获取原节点的next在新节点中的位置，以及用hash[cur]->next对当前的新节点的指针赋值

Node* copyRandomList(Node* head) {
    if (!head) return head;
    unordered_map<Node*, Node*> hash;
    Node *cur = head;
    while (cur) {   // 第一次遍历只生成节点
        Node *tmp = new Node(cur->val);
        hash[cur] = tmp;
        cur = cur->next;
    }
    cur = head;
    while (cur) {   // 第二次遍历对指针赋值
        if (cur->next) {
            hash[cur]->next = hash[cur->next];
        }
        if (cur->random) {
            hash[cur]->random = hash[cur->random];
        }
        cur = cur->next;
    }
    return hash[head];
}