138. Copy List with Random Pointer

source: https://leetcode.com/problems/copy-list-with-random-pointer/

Question

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.

For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

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.

思路

这道题的思路比较直球，用一个数组存储链表的所有元素，然后用引用地址找到index，再复制。

/**
 * @param {Node} head
 * @return {Node}
 */
var copyRandomList = function (head) {
    if (!head) return head;
    const arr = [];
    let crr = head;
    while (crr) {
        arr.push(crr);
        crr = crr.next;
    }
    const list = arr.map(node => new Node(node.val, null, null));
    arr.forEach((node, i) => {
        if (node.random) {
            const randomIndex = arr.findIndex(n => node.random === n);
            list[i].random = list[randomIndex];
        }
        list[i].next = list[i + 1] || null;
    });
    return list[0];
};

总结

对于比较直接的思路，有足够高的熟练度就可以做出。