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[Leetcode]382. Linked List Random Node(C++)

题目描述

题目链接: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.

例子

例子 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.

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?

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.

解题思路

这道题思路比较直观,首先记录链表的长度,每次随机去一个数字取对应的节点的值即可。

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class Solution {
public:
/** @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(ListNode* head) : head_(head) {
ListNode* curr = head;
length_ = 0;
while (curr) {
length_++;
curr = curr->next;
}
}

/** Returns a random node's value. */
int getRandom() {
int step = rand() % length_;
ListNode* ret = head_;
while (step-- != 0) {
ret = ret->next;
}
return ret->val;
}

private:
ListNode* head_;
int length_;
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(n)

GitHub 代码同步地址: 382.LinkedListRandomNode.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions