Given the head
of a singly linked list, return true
if it is a
or false
otherwise.
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?
A:
就是简单的读2遍,然后一个从后往前,一个从前往后。
/**
* 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) {
vector<int> v;
ListNode* runner = head;
while(runner){
v.push_back(runner->val);
runner = runner->next;
}
runner = head;
while(runner){
int val = v.back();
v.pop_back();
if(val != runner->val){
return false;
}
runner = runner->next;
}
return true;
}
};
Follow up:
Could you do it in O(n)
time and O(1)
space?
1: reverse后半段。
2: 利用 recursive 的call stack,来对比