【LBLD】如何判断回文链表

判断回文单链表

/**
 * 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:
    ListNode *left;

    bool isPalindrome(ListNode* head) {
        left = head;
        return traverse(head);
    }

    bool traverse(ListNode* head) {
        if (head == nullptr) return true;
        bool res = traverse(head->next);
        res = res && (left->val == head->val);
        left = left->next;
        return res;
    }
};

优化空间复杂度

使用快慢指针找到链表中点，然后把链表后半段反转，双指针判断两个链表是否相同。

/**
 * 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) {
        // 使用快慢指针找到中点
        ListNode *slow = head, *fast = head;
        while (nullptr != fast && nullptr != fast->next) {
            fast = fast->next->next;
            slow = slow->next;
        }
        if (nullptr != fast) {
            slow = slow->next;
        }

        ListNode *left = head;
        ListNode *right = reverse(slow);
        while (nullptr != right) {
            if (right->val != left->val) return false;
            left = left->next;
            right = right->next;
        }
        return true;
    }

    ListNode* reverse(ListNode* head) {
        ListNode *pre = nullptr;
        ListNode *cur = head;
        ListNode *nxt = head;

        while (cur) {
            nxt = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nxt;
        }
        return pre;
    }
};