382. 链表随机节点

题目描述

给你一个单链表，随机选择链表的一个节点，并返回相应的节点值。每个节点 被选中的概率一样 。

实现 Solution 类：

Solution(ListNode head) 使用整数数组初始化对象。
int getRandom() 从链表中随机选择一个节点并返回该节点的值。链表中所有节点被选中的概率相等。

示例：

输入
["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
[[[1, 2, 3]], [], [], [], [], []]
输出
[null, 1, 3, 2, 2, 3]

解释
Solution solution = new Solution([1, 2, 3]);
solution.getRandom(); // 返回 1
solution.getRandom(); // 返回 3
solution.getRandom(); // 返回 2
solution.getRandom(); // 返回 2
solution.getRandom(); // 返回 3
// getRandom() 方法应随机返回 1、2、3中的一个，每个元素被返回的概率相等。

提示：

链表中的节点数在范围 [1, 10⁴] 内
-10⁴ <= Node.val <= 10⁴
至多调用 getRandom 方法 10⁴ 次

进阶：

如果链表非常大且长度未知，该怎么处理？
你能否在不使用额外空间的情况下解决此问题？

解法

方法一

Python3JavaC++Go

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def __init__(self, head: Optional[ListNode]):
        self.head = head

    def getRandom(self) -> int:
        n = ans = 0
        head = self.head
        while head:
            n += 1
            x = random.randint(1, n)
            if n == x:
                ans = head.val
            head = head.next
        return ans


# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    private ListNode head;
    private Random random = new Random();

    public Solution(ListNode head) {
        this.head = head;
    }

    public int getRandom() {
        int ans = 0, n = 0;
        for (ListNode node = head; node != null; node = node.next) {
            ++n;
            int x = 1 + random.nextInt(n);
            if (n == x) {
                ans = node.val;
            }
        }
        return ans;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */

/**
 * 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* head;

    Solution(ListNode* head) {
        this->head = head;
    }

    int getRandom() {
        int n = 0, ans = 0;
        for (ListNode* node = head; node != nullptr; node = node->next) {
            n += 1;
            int x = 1 + rand() % n;
            if (n == x) ans = node->val;
        }
        return ans;
    }
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution* obj = new Solution(head);
 * int param_1 = obj->getRandom();
 */

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
type Solution struct {
    head *ListNode
}

func Constructor(head *ListNode) Solution {
    return Solution{head}
}

func (this *Solution) GetRandom() int {
    n, ans := 0, 0
    for node := this.head; node != nil; node = node.Next {
        n++
        x := 1 + rand.Intn(n)
        if n == x {
            ans = node.Val
        }
    }
    return ans
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(head);
 * param_1 := obj.GetRandom();
 */

382. 链表随机节点

题目描述

解法

方法一

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