Linked List Random Node Problem
Description
LeetCode Problem 382.
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:
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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, 10^4].
- -10^4 <= Node.val <= 10^4
- At most 10^4 calls will be made to getRandom.
Sample C++ Code
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* HeadNode;
Solution(ListNode* head) {
HeadNode = head;
}
int getRandom() {
int res, len = 1;
ListNode* x = HeadNode;
while(x){
if(rand() % len == 0){
res = x->val;
}
len++;
x = x->next;
}
return res;
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution* obj = new Solution(head);
* int param_1 = obj->getRandom();
*/