Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2 Output: false
Example 2:
Input: 1->2->2->1 Output: true
Follow up:
Could you do it in O(n) time and O(1) space?
解题:
找到中间节点后反转判断节点回文
实现:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
bool isPalindrome(struct ListNode* head){
if (head == NULL || head->next == NULL) {
return true;
}
struct ListNode* slow=head;
struct ListNode* fast=head;
struct ListNode* pre=NULL;
struct ListNode* next=NULL;
//get middle
while( fast!=NULL && fast->next!=NULL){
slow = slow->next;
fast = fast->next->next;
}
if(fast) {
slow=slow->next;
}
struct ListNode* cur=slow;
//and reverse
while(cur!=NULL){
pre = cur;
cur = cur->next;
pre->next = next;
next = pre;
}
//compare
cur = head;
while(pre!=NULL){
//printf("%d %d\n", head->val, pre->val);
if (cur->val != pre->val){
return false;
}
cur=cur->next;
pre= pre->next;
}
return true;
}
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