# 234. Palindrome Linked List (Easy)

## 1.問題

* 判斷linked list是否屬於迴文

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LI-SBaOFox_JpsaAXbA%2F-LI-Sces7uTu3trUmtCo%2F2018072208.jpg?alt=media\&token=e3f10bee-30b8-4ab0-bcbb-8e72a9c996d1)

## 2.想法 

## 3.程式碼

```
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        string str;
        ListNode *list = head;
        while(list) {
            str.push_back(list->val);
            list = list->next;
        }
        for (int i = 0; i < str.size(); i++) {
            if (str[i] != str[str.size() - 1 - i]){
                return false;
            }
        }
        return true;
    }
};
```

## 4.Performance

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LI-SBaOFox_JpsaAXbA%2F-LI-T3huilKxnc8OUFA9%2F2018072209.jpg?alt=media\&token=98d4c844-d5bd-42b3-af19-1cc85d5a859c)