leetcode 234. 回文链表

2023.9.5

        本题先将链表的节点值移到数组中，再用双指针去判断该数组是否为回文的即可。 代码如下：

/**
 * 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> nums;
        while(head)
        {
            nums.push_back(head->val);
            head = head->next;
        }
        int left = 0;
        int right = nums.size()-1;
        while(left <= right)
        {
            if(nums[left] != nums[right]) return false;
            left++;
            right--;
        }
        return true;
    }
};

2023.11.10

        二刷，思路还是先将链表的值放入数组中，判断数组是否回文就简单多了。 java代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
         List nums = new ArrayList<>();
         nums.add(head.val);
         while(head.next != null){
             head = head.next;
             nums.add(head.val);
         }
         for(int i=0; i
             if(nums.get(i) != nums.get(nums.size()-i-1)) return false;
         }
         return true;
    }
}