LeetCode-in-Cpp
234. Palindrome Linked List
Easy
Given the head of a singly linked list, return true if it is a palindrome.
Example 1:
Input: head = [1,2,2,1]
Output: true
Example 2:
Input: head = [1,2]
Output: false
Constraints:
- The number of nodes in the list is in the range
[1, 105]. 0 <= Node.val <= 9
Follow up: Could you do it in O(n) time and O(1) space?
Solution
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
bool isPalindrome(ListNode* head) {
int len = 0;
ListNode* right = head;
// Calculate the length
while (right != nullptr) {
right = right->next;
len++;
}
// Reverse the right half of the list
len = len / 2;
right = head;
for (int i = 0; i < len; i++) {
right = right->next;
}
ListNode* prev = nullptr;
while (right != nullptr) {
ListNode* next = right->next;
right->next = prev;
prev = right;
right = next;
}
// Compare left half and right half
for (int i = 0; i < len; i++) {
if (prev != nullptr && head->val == prev->val) {
head = head->next;
prev = prev->next;
} else {
return false;
}
}
return true;
}
};