382. Linked List Random Node: 隨機選元素從list裡面

 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 integer array nums.
• 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 choosen.

 

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?

這題是reservoir computing

很難

沒讀過就不會

class Solution:

    def __init__(self, head: Optional[ListNode]):
        self.head=head

    def getRandom(self) -> int:
        
        scope=1
        curr=self.head
        chosen_value = 0
        while curr:
            # decide whether to include the element in reservoir
            if random.random() < 1 / scope:
                chosen_value = curr.val
            # move on to the next node
            curr = curr.next
            scope += 1
        return chosen_value