57-Palindrome Linked List

1. Palindrome Linked List My Submissions QuestionEditorial Solution
   Total Accepted: 46990 Total Submissions: 166743 Difficulty: Easy
   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?

先给出一种时间O(n)空间O(n)的算法,思路就是顺序把值放入到顺序表
然后在顺序表判定回文

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

        while(head!=NULL){
            vals.push_back(head->val);
            head = head->next;
        }
        int n = vals.size(),flag=0;
        for(int i=0;i<n/2;++i)
            if(vals[i]!=vals[n-1-i]){
                flag =1;
                break;
            }
        if(flag==1)return false;
        else return true;
    }
};

给出O(n)空间O(1)的算法，虽然代码长很多，但是省空间啊！！
思想：O->O->O->O->O前半部分逆转，然后中间出发向两端比较

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if(head==NULL||head->next==NULL)return true;
        int len=0;
        ListNode *p =head;
        while(p!=NULL){
            len++;
            p=p->next;
        }
        if(len==2)return head->next->val==head->val; //因为逆转的时候是针对大于等于2（2*2就是4）的情况来考虑，所以先处理小于4 
        if(len==3)return head->val==head->next->next->val;
        int i=1;
        ListNode *fir=head,*sec=head;
        while(i++<len/2)fir=fir->next;
        if(len%2==0)sec = fir->next;
        else sec = fir->next->next;
        if(len>=4){          //逆转前半部分 
            ListNode *pre_next = fir->next;
            ListNode *tmp=head->next,*pretmp = head;
            while(tmp!=NULL&&tmp!=pre_next){ 
                ListNode *tmpnext = tmp->next;
                tmp->next = pretmp;
                pretmp = tmp;
                tmp=tmpnext;
            }
            head->next=NULL;
        }
        while(fir!=NULL){  //从中间出发判断回文 
            if(fir->val!=sec->val)return false;
            fir = fir->next;
            sec = sec->next;
        }
        return true;
    }
};