https://leetcode.com/explore/challenge/card/30-day-leetcoding-challenge/529/week-2/3290/
>> Solution 1 : Find Size, Then Middle
O(n) time, O(1) space--
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/** | |
* Definition for singly-linked list. | |
* public class ListNode { | |
* int val; | |
* ListNode next; | |
* ListNode(int x) { val = x; } | |
* } | |
*/ | |
class LinkedListMiddleNode { | |
public ListNode solve(ListNode head) { | |
int size = 0; | |
// Go to the last node to find the size | |
ListNode current = head; | |
while(current != null) { | |
current = current.next; | |
size++; | |
} | |
int middle = size/2; | |
// Go to the middle node | |
current = head; | |
for(int i=0; i < middle; i++) { | |
current = current.next; | |
} | |
return current; | |
} | |
} |
>> Solution 2 : Slow and Fast Pointer
O(n) time, O(1) space--
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/** | |
* Definition for singly-linked list. | |
* public class ListNode { | |
* int val; | |
* ListNode next; | |
* ListNode(int x) { val = x; } | |
* } | |
*/ | |
class LinkedListMiddleNodeFastPointer { | |
public ListNode middleNode(ListNode head) { | |
ListNode sizePointer = head; | |
ListNode middlePointer = head; | |
while(sizePointer != null && sizePointer.next != null) { | |
middlePointer = middlePointer.next; | |
sizePointer = sizePointer.next.next; | |
} | |
return middlePointer; | |
} | |
} |
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