382. Linked List Random Node

Medium

Given a singly linked list, return a random node's value from the linked list. Each node must have the same probability of being chosen.

Implement the Solution class:

• Solution(ListNode head) Initializes the object with the head of the singly-linked list head.

• int getRandom() Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be chosen.

Example 1:

Input
["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 3, 2, 2, 3]

Explanation
Solution solution = new Solution([1, 2, 3]);
solution.getRandom(); // return 1
solution.getRandom(); // return 3
solution.getRandom(); // return 2
solution.getRandom(); // return 2
solution.getRandom(); // return 3
// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.

Constraints:

• The number of nodes in the linked list will be in the range [1, 104].

• -104 <= Node.val <= 104

• At most 104 calls will be made to getRandom.

Follow up:

• What if the linked list is extremely large and its length is unknown to you?

• Could you solve this efficiently without using extra space?

解題

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
type Solution struct {
    arr []int
}


func Constructor(head *ListNode) Solution {
    arr := make([]int, 0)
    for head != nil {
        arr = append(arr, head.Val)
        head = head.Next
    }

    return Solution{ arr: arr }
}


func (this *Solution) GetRandom() int {
    return this.arr[rand.Intn(len(this.arr))]
}


/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(head);
 * param_1 := obj.GetRandom();
 */