234. Palindrome Linked List

找到中点reverse后半部，跟前半部比对即可，比如[1,2,1]拆成[1]和[2,1]，比较[1]跟[1,2]即可

/**
 * 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) {
        if (!head || !head->next) return true;
        ListNode dummy_head(0), *slow = &dummy_head, *fast = head; // fast从head或者dummy_head开始都行
        dummy_head.next = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = slow->next;
        slow->next = nullptr;
        fast = reverse(fast);
        while (head && head->val == fast->val) { // head短fast长因为找中点时fast从head开始
            head = head->next;
            fast = fast->next;
        }
        return !head;
    }

    ListNode *reverse(ListNode *head) {
        ListNode *prev = nullptr, *curr = head;
        while (curr) {
            auto succ = curr->next; curr->next = prev; prev = curr;
            curr = succ;
        }
        return prev;
    }
};

/**
 * 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) {
        if (!head || !head->next) return true;
        ListNode *prev = nullptr, *slow = head, *fast = head;
        while (fast && fast->next) {
            prev = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = fast ? slow->next : slow;
        auto l2 = reverse(fast);
        prev->next = nullptr;
        while (head && l2 && head->val == l2->val) {
            head = head->next;
            l2 = l2->next;
        }
        return !head;
    }

    ListNode *reverse(ListNode *head) {
        ListNode *prev = nullptr, *curr = head;
        while (curr) {
            auto succ = curr->next;
            curr->next = prev;
            prev = curr;
            curr = succ;
        }
        return prev;
    }
};

/**
 * 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) {
        if (!head || !head->next) return true;
        ListNode *prev = nullptr, *slow = head, *fast = head;
        while (fast && fast->next) {
            prev = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = fast ? slow->next : slow;
        auto l2 = reverse(fast);
        prev->next = nullptr;
        while (head && l2 && head->val == l2->val) {
            head = head->next;
            l2 = l2->next;
        }
        return !head;
    }

    ListNode *reverse(ListNode *head) {
        if (!head || !head->next) return head;
        auto res = reverse(head->next);
        head->next->next = head;
        head->next = nullptr;
        return res;
    }
};

/**
 * 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) {
        if (!head || !head->next) return true;
        auto slow = head, fast = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = slow->next;
        auto l2 = reverse(fast);
        slow->next = nullptr;
        if (!fast->next) {
            fast->next = slow;
        }
        while (head && l2 && head->val == l2->val) {
            head = head->next;
            l2 = l2->next;
        }
        return !head;
    }

    ListNode *reverse(ListNode *head) {
        if (!head || !head->next) return head;
        auto res = reverse(head->next);
        head->next->next = head;
        head->next = nullptr;
        return res;
    }
};