Linked List Random Node
2016-08-20 13:57
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Given a singly linked list, return a random node's value from the linked list. Each node must have thesame probability of being chosen.
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?
Example:
思路:用hashmap来存node,和index,然后用随机函数来取。
Follow up question: 实际上就是水塘采样。算法就是随时get当前的长度,取random [0,n),如果取到0就取当前刚进来的node,如果不是,那就取之前那个random的值,这样能保证每个点都是随机的。证明过程,有个状态图。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
/** @param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node. */
private ListNode curNode = null;
private ListNode preNode = null;
private ListNode beginNode = null;
public Solution(ListNode head) {
beginNode = head;
}
/** Returns a random node's value. */
public int getRandom() {
preNode = beginNode;
curNode = beginNode;
int count = 1;
while(curNode != null) {
int randValue = (int)(Math.random()*count);
if(randValue == 0){
preNode = curNode;
}
curNode = curNode.next;
count++;
}
return preNode.val;
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(head);
* int param_1 = obj.getRandom();
*/
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?
Example:
// Init a singly linked list [1,2,3]. ListNode head = new ListNode(1); head.next = new ListNode(2); head.next.next = new ListNode(3); Solution solution = new Solution(head); // getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning. solution.getRandom();
思路:用hashmap来存node,和index,然后用随机函数来取。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { /** @param head The linked list's head. Note that the head is guaranteed to be not null, so it contains at least one node. */ HashMap<Integer, ListNode> hashmap = new HashMap<Integer, ListNode>(); private int length = 0; public Solution(ListNode head) { ListNode node = head; int index = 1; while(node!=null){ length++; hashmap.put(index, node); node = node.next; index++; } } /** Returns a random node's value. */ public int getRandom() { int value = (int)(Math.random()*length)+1; return hashmap.get(value).val; } } /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(head); * int param_1 = obj.getRandom(); */
Follow up question: 实际上就是水塘采样。算法就是随时get当前的长度,取random [0,n),如果取到0就取当前刚进来的node,如果不是,那就取之前那个random的值,这样能保证每个点都是随机的。证明过程,有个状态图。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
/** @param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node. */
private ListNode curNode = null;
private ListNode preNode = null;
private ListNode beginNode = null;
public Solution(ListNode head) {
beginNode = head;
}
/** Returns a random node's value. */
public int getRandom() {
preNode = beginNode;
curNode = beginNode;
int count = 1;
while(curNode != null) {
int randValue = (int)(Math.random()*count);
if(randValue == 0){
preNode = curNode;
}
curNode = curNode.next;
count++;
}
return preNode.val;
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(head);
* int param_1 = obj.getRandom();
*/
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