LeetCode //C - 382. Linked List Random Node

382. Linked List Random Node

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 , 1 0 4 ] [1, 10^4] [1,104].
  • − 1 0 4 < = N o d e . v a l < = 1 0 4 -10^4 <= Node.val <= 10^4 −104<=Node.val<=104
  • At most 1 0 4 10^4 104 calls will be made to getRandom.

From: LeetCode

Link: 382. Linked List Random Node


Solution:

Ideas:
  1. solutionCreate: Initializes the Solution object, stores the head of the linked list, and seeds the random number generator.

  2. solutionGetRandom: Implements reservoir sampling to randomly select a node with equal probability. It iterates over the entire linked list and replaces the current result with a random chance, ensuring each node has an equal chance of being picked.

  3. solutionFree: Frees the memory allocated for the Solution object.

Code:
c 复制代码
// Definition for singly-linked list.
struct {
    int val;
    struct ListNode *next;
}ListNode;

typedef struct {
    struct ListNode* head;
} Solution;

Solution* solutionCreate(struct ListNode* head) {
    Solution* obj = (Solution*) malloc(sizeof(Solution));
    obj->head = head;
    srand(time(NULL)); // Initialize random number generator
    return obj;
}

int solutionGetRandom(Solution* obj) {
    struct ListNode* current = obj->head;
    int result = current->val;
    int i = 1;
    
    while (current != NULL) {
        // Pick a random number in range [0, i]
        if (rand() % i == 0) {
            result = current->val; // Replace the result with current node's value
        }
        current = current->next;
        i++;
    }
    return result;
}

void solutionFree(Solution* obj) {
    free(obj);
}

/**
 * Your Solution struct will be instantiated and called as such:
 * Solution* obj = solutionCreate(head);
 * int param_1 = solutionGetRandom(obj);
 * solutionFree(obj);
 */
相关推荐
地平线开发者9 小时前
SparseDrive 模型导出与性能优化实战
算法·自动驾驶
董董灿是个攻城狮10 小时前
大模型连载2:初步认识 tokenizer 的过程
算法
地平线开发者10 小时前
地平线 VP 接口工程实践(一):hbVPRoiResize 接口功能、使用约束与典型问题总结
算法·自动驾驶
罗西的思考10 小时前
AI Agent框架探秘:拆解 OpenHands(10)--- Runtime
人工智能·算法·机器学习
HXhlx14 小时前
CART决策树基本原理
算法·机器学习
Wect14 小时前
LeetCode 210. 课程表 II 题解:Kahn算法+DFS 双解法精讲
前端·算法·typescript
颜酱15 小时前
单调队列:滑动窗口极值问题的最优解(通用模板版)
javascript·后端·算法
Gorway21 小时前
解析残差网络 (ResNet)
算法
拖拉斯旋风21 小时前
LeetCode 经典算法题解析:优先队列与广度优先搜索的巧妙应用
算法
Wect21 小时前
LeetCode 207. 课程表:两种解法(BFS+DFS)详细解析
前端·算法·typescript