leetcode/0382.Linked-List-Random-Node/README.md
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 integer array nums.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 choosen.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:
[1, 104].-10^4 <= Node.val <= 10^410^4 calls will be made to getRandom.Follow up:
给定一个单链表,随机选择链表的一个节点,并返回相应的节点值。保证每个节点被选的概率一样。
进阶: 如果链表十分大且长度未知,如何解决这个问题?你能否使用常数级空间复杂度实现?
package leetcode
import (
"math/rand"
"github.com/halfrost/LeetCode-Go/structures"
)
// ListNode define
type ListNode = structures.ListNode
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
type Solution struct {
head *ListNode
}
/** @param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node. */
func Constructor(head *ListNode) Solution {
return Solution{head: head}
}
/** Returns a random node's value. */
func (this *Solution) GetRandom() int {
scope, selectPoint, curr := 1, 0, this.head
for curr != nil {
if rand.Float64() < 1.0/float64(scope) {
selectPoint = curr.Val
}
scope += 1
curr = curr.Next
}
return selectPoint
}
/**
* Your Solution object will be instantiated and called as such:
* obj := Constructor(head);
* param_1 := obj.GetRandom();
*/