刷题--回文链表

1.暴力解法

/**
 * 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) {}
 * };
 */
#include <vector>
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode* tempnode = head;
        std::vector<int> val_list;
        while (tempnode != nullptr) {
            val_list.push_back(tempnode->val);
            tempnode = tempnode->next;
        }
        for (int i = 0; i < val_list.size(); i++) {
            if (val_list[i] != val_list[val_list.size() - (i + 1)])
                return false;
        }
        return true;
    }
};

2.暴力解法-双指针法

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        std::vector<int> vals;

        // Step 1: 把链表值复制进数组
        while (head) {
            vals.push_back(head->val);
            head = head->next;
        }

        // Step 2: 双指针判断是否是回文
        int left = 0, right = vals.size() - 1;
        while (left < right) {
            if (vals[left] != vals[right]) return false;
            ++left;
            --right;
        }

        return true;
    }
};