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
104
calls 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?
SOLUTION:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
import random
class Solution:
def __init__(self, head: Optional[ListNode]):
self.head = head
def getRandom(self) -> int:
res = self.head.val
n = 2
curr = self.head.next
while curr:
isSwap = random.randrange(0, n)
if isSwap == 0:
res = curr.val
curr = curr.next
n += 1
return res
# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()
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