VietMX Staff asked 3 years ago
Problem
Could you do it in O(1) time and O(1) space?
Solution: is Reversed first half == Second half?
Let’s look to an example [1, 1, 2, 1]. In the beginning, set two pointers fast and slow starting at the head.
1 -> 1 -> 2 -> 1 -> null
sf(1) Move: fast pointer goes to the end, and slow goes to the middle.
1 -> 1 -> 2 -> 1 -> null
s f(2) Reverse: the right half is reversed, and slow pointer becomes the 2nd head.
1 -> 1 null <- 2 <- 1
h s(3) Compare: run the two pointers head and slow together and compare.
1 -> 1 null <- 2 <- 1
h sImplementation
public boolean isPalindrome(ListNode head) {
ListNode fast = head, slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
if (fast != null) { // odd nodes: let right half smaller
slow = slow.next;
}
slow = reverse(slow);
fast = head;
while (slow != null) {
if (fast.val != slow.val) {
return false;
}
fast = fast.next;
slow = slow.next;
}
return true;
}
public ListNode reverse(ListNode head) {
ListNode prev = null;
while (head != null) {
ListNode next = head.next;
head.next = prev;
prev = head;
head = next;
}
return prev;
}