Kth Last Node of a Linked List - Google Top Interview Questions
Problem Statement :
Given a singly linked list node, return the value of the kth last node (0-indexed).
k is guaranteed not to be larger than the size of the linked list.
This should be done in \mathcal{O}(1)O(1) space.
Constraints
n ≤ 100,000 where n is the length of node
Example 1
Input
node = [1, 2]
k = 1
Output
1
Explanation
The second last node has the val of 1
Example 2
Input
node = [0, 1, 2, 3]
k = 2
Output
1
Solution :
Solution in C++ :
int solve(LLNode* node, int k) {
LLNode* tmp = node;
int i = 0;
while (tmp != NULL) {
if (i > k) {
node = node->next;
}
i++;
tmp = tmp->next;
}
return node->val;
}
Solution in Java :
import java.util.*;
/**
* class LLNode {
* int val;
* LLNode next;
* }
*/
class Solution {
public int solve(LLNode node, int k) {
int size = 0;
LLNode head = node;
while (head != null) {
size++;
head = head.next;
}
LLNode curr = node;
while (curr != null) {
size--;
if (size == k)
return curr.val;
curr = curr.next;
}
return 0;
}
}
Solution in Python :
class Solution:
def solve(self, node, k):
result, end = node, node
count = 0
while end:
if count > k:
result = result.next
end = end.next
count += 1
return result.val if count > k else None
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