leetcode：Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

分析：题意为判断单链表是否为回文的。

思路：首先想到的是 遍历一次单链表，将其元素装入vector，然后进行第二次遍历比较来判断回文。

/**
 * 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) {
        vector<int> temp;
        while(head){
            temp.push_back(head->val);
            head=head->next;
        }
        for(int i=0,j=temp.size()-1;i<j;i++,j--){
            if(temp[i]!=temp[j]) return false;
        }
        return true;
    }
};

　可是这种方法：时间复杂度O(n)，空间复杂度O(n)；

为了使空间复杂度为O(1)，可以不采用vector等，思路：找到单链表的中点，进行拆分，逆转后半个链表，然后对这两个链表同时顺序遍历一次进行判断。　

/**
 * 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 == NULL || head->next == NULL)
            return true;
        ListNode* mid = getMid(head);
        ListNode* head2 = reverse(mid);
        while(head && head2)
        {
            if(head->val != head2->val)
                return false;
            head = head->next;
            head2 = head2->next;
        }
        return true;
    }
    ListNode* getMid(ListNode* head)
    {// at least two nodes
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* preslow = NULL;
        do
        {
            fast = fast->next;
            if(fast)
            {
                fast = fast->next;
                preslow = slow;
                slow = slow->next;
            }
        }while(fast != NULL);
        preslow->next = NULL;
        return slow;
    }
    ListNode* reverse(ListNode* head)
    {
        if(head == NULL || head->next == NULL)
            return head;
        else if(head->next->next == NULL)
        {// two nodes
            ListNode* tail = head->next;
            head->next = NULL;
            tail->next = head;
            return tail;
        }
        else
        {
            ListNode* pre = head;
            ListNode* cur = pre->next;
            pre->next = NULL;   // set tail
            ListNode* post = cur->next;
            while(post)
            {
                cur->next = pre;
                pre = cur;
                cur = post;
                post = post->next;
            }
            cur->next = pre;
            return cur;
        }
    }
};

　或者：

同样O(n) time and O(1) space c++, fast and slow pointer

class Solution {
public:
    bool isPalindrome(ListNode* slow, ListNode* fast)
    {
      if (fast == nullptr) {
        half = slow;
        return true;
      }
      if (fast->next == nullptr) {
        half = slow->next;
        return true;
      }

      if (isPalindrome(slow->next, fast->next->next) && slow->val == half->val) {
        half = half->next;
        return true;
      }

      return false;
    }

    bool isPalindrome(ListNode* head) {
      return isPalindrome(head, head);
    }

    ListNode* half;
};