234回文链表

 分块写逻辑更清楚一点

思路：快慢指针查找中点，然后翻转链表206题，然后比较是否相等

时间复杂度：3/2 O(n) ,空间复杂度O(1)

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if (head == nullptr) {
            return true;
        }

        // 找到前半部分链表的尾节点并反转后半部分链表
        ListNode* firstHalfEnd = endOfFirstHalf(head);
        ListNode* secondHalfStart = reverseList(firstHalfEnd->next);

        // 判断是否回文
        ListNode* p1 = head;
        ListNode* p2 = secondHalfStart;
        bool result = true;
        while (result && p2 != nullptr) {
            if (p1->val != p2->val) {
                result = false;
            }
            p1 = p1->next;
            p2 = p2->next;
        }        

        // 还原链表并返回结果
        firstHalfEnd->next = reverseList(secondHalfStart);
        return result;
    }

    ListNode* reverseList(ListNode* head) {
        ListNode* prev = nullptr;
        ListNode* curr = head;
        while (curr != nullptr) {
            ListNode* nextTemp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }

    ListNode* endOfFirstHalf(ListNode* head) {
        ListNode* fast = head;
        ListNode* slow = head;
        while (fast->next != nullptr && fast->next->next != nullptr) {
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }
};

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode*fast=head;
        ListNode*slow=head;
        while(fast!=nullptr&&fast->next!=nullptr){
            fast=fast->next->next;
            slow=slow->next;
        }
        ListNode*pre=nullptr;
        ListNode*cur=slow;
        while(cur!=nullptr){
            ListNode*temp=cur->next;
            cur->next=pre;
            pre=cur;
            cur=temp;
        }
        while(pre!=nullptr){
            if(pre->val!=head->val)
                return false;
            pre=pre->next;
            head=head->next;
        }
        return true;
    }
};

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        vector<ListNode*>vec;
        while(head!=nullptr){
            vec.emplace_back(head);
            head=head->next;
        }
        for(int i=0,j=vec.size()-1;i<j;i++,j--){
            if(vec[i]->val!=vec[j]->val)
                return false;
        }
        return true;
    }
};

递归

相当于在每次递归中判断当前递归函数中和temp是否相等，相等返回true递归到上一次curNode中，不相等就false

class Solution {
public:
    ListNode*temp;
    bool myPalind(ListNode*curNode){
        if(curNode!=nullptr){
            if(!myPalind(curNode->next))
                return false;
             if(curNode->val!=temp->val)
                 return false;
             temp=temp->next;
        }
        return true;
    }
    bool isPalindrome(ListNode* head) {
        temp=head;
        return myPalind(head);
    }
};