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Add solution 0382
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package leetcode
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import (
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"math/rand"
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"github.com/halfrost/LeetCode-Go/structures"
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)
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// ListNode define
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type ListNode = structures.ListNode
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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type Solution struct {
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head *ListNode
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}
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/** @param head The linked list's head.
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Note that the head is guaranteed to be not null, so it contains at least one node. */
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func Constructor(head *ListNode) Solution {
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return Solution{head: head}
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}
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/** Returns a random node's value. */
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func (this *Solution) GetRandom() int {
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scope, selectPoint, curr := 1, 0, this.head
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for curr != nil {
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if rand.Float64() < 1.0/float64(scope) {
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selectPoint = curr.Val
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}
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scope += 1
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curr = curr.Next
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}
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return selectPoint
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}
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/**
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* Your Solution object will be instantiated and called as such:
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* obj := Constructor(head);
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* param_1 := obj.GetRandom();
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*/
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package leetcode
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import (
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"fmt"
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"testing"
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"github.com/halfrost/LeetCode-Go/structures"
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)
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func Test_Problem382(t *testing.T) {
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header := structures.Ints2List([]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 0})
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obj := Constructor(header)
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fmt.Printf("obj = %v\n", structures.List2Ints(header))
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param1 := obj.GetRandom()
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fmt.Printf("param_1 = %v\n", param1)
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param1 = obj.GetRandom()
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fmt.Printf("param_1 = %v\n", param1)
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param1 = obj.GetRandom()
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fmt.Printf("param_1 = %v\n", param1)
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param1 = obj.GetRandom()
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fmt.Printf("param_1 = %v\n", param1)
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param1 = obj.GetRandom()
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fmt.Printf("param_1 = %v\n", param1)
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param1 = obj.GetRandom()
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fmt.Printf("param_1 = %v\n", param1)
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fmt.Printf("\n\n\n")
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}
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105
leetcode/0382.Linked-List-Random-Node/README.md
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105
leetcode/0382.Linked-List-Random-Node/README.md
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# [382. Linked List Random Node](https://leetcode.com/problems/linked-list-random-node/)
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## 题目
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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.
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Implement the `Solution` class:
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- `Solution(ListNode head)` Initializes the object with the integer array nums.
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- `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.
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**Example 1:**
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```
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Input
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["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
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[[[1, 2, 3]], [], [], [], [], []]
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Output
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[null, 1, 3, 2, 2, 3]
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Explanation
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Solution solution = new Solution([1, 2, 3]);
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solution.getRandom(); // return 1
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solution.getRandom(); // return 3
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solution.getRandom(); // return 2
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solution.getRandom(); // return 2
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solution.getRandom(); // return 3
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// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
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```
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**Constraints:**
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- The number of nodes in the linked list will be in the range `[1, 104]`.
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- `-10^4 <= Node.val <= 10^4`
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- At most `10^4` calls will be made to `getRandom`.
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**Follow up:**
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- What if the linked list is extremely large and its length is unknown to you?
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- Could you solve this efficiently without using extra space?
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## 题目大意
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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。
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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?
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## 解题思路
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- rand.Float64() 可以返回 [0.0,1.0) 之间的随机数。利用这个函数完成我们的随机化取节点的过程。
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## 代码
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```go
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package leetcode
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import (
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"math/rand"
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"github.com/halfrost/LeetCode-Go/structures"
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)
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// ListNode define
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type ListNode = structures.ListNode
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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type Solution struct {
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head *ListNode
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}
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/** @param head The linked list's head.
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Note that the head is guaranteed to be not null, so it contains at least one node. */
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func Constructor(head *ListNode) Solution {
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return Solution{head: head}
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}
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/** Returns a random node's value. */
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func (this *Solution) GetRandom() int {
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scope, selectPoint, curr := 1, 0, this.head
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for curr != nil {
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if rand.Float64() < 1.0/float64(scope) {
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selectPoint = curr.Val
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}
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scope += 1
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curr = curr.Next
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}
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return selectPoint
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}
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/**
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* Your Solution object will be instantiated and called as such:
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* obj := Constructor(head);
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* param_1 := obj.GetRandom();
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*/
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```
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