leetcode 382. Linked List Random Node_leetcode 382 c语言-优快云博客

本文链接：https://blog.youkuaiyun.com/zhaihaixu/article/details/78684093

本文介绍了一个从单链表中返回一个随机节点值的算法实现，该算法确保每个节点被选中的概率相同。此外，还讨论了如何高效地处理非常大的链表而不需要额外的空间。

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原题：

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.

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?

Example:

// Init a singly linked list [1,2,3].
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
Solution solution = new Solution(head);

// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
solution.getRandom();

代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
typedef struct {
    int* nums;
    int len;
} Solution;

/** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
Solution* solutionCreate(struct ListNode* head) {
    Solution *obj;
    obj=(Solution*)malloc(sizeof(Solution));
    obj->nums=(int*)malloc(sizeof(int)*10000);
    struct ListNode* temp;
    temp=head;
    int n=0;
    while(temp!=NULL)
    {
        *(obj->nums+n)=temp->val;
        n++;
        temp=temp->next;
    }
    obj->len=n;
    return obj;
}

/** Returns a random node's value. */
int solutionGetRandom(Solution* obj) {
    clock_t t = clock();
    int num=t%obj->len;
    return *(obj->nums+num);
}

void solutionFree(Solution* obj) {
    free(obj->nums);
    free(obj);
}

/**
 * Your Solution struct will be instantiated and called as such:
 * struct Solution* obj = solutionCreate(head);
 * int param_1 = solutionGetRandom(obj);
 * solutionFree(obj);
 */

话说代码很简单，但是我一直不明白这个验证是怎么验证的。

就因为代码对比不对？还是因为什么别的。。

我再看看。