Question

382. 链表随机节点

English Version

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

# 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();
 */