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 listhead.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
104calls will be made togetRandom.
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?
/**
* 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();
*/
Last modified 5mo ago