Palindrome Linked List

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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?

判断是不是回文链表

<p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;"></p>/** * 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;            }        }    }};<p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;"></p>

/** * 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;            }        }    }};

/** * 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;       stack<int> temp;       ListNode* cur=head;       while(cur){           temp.push(cur->val);           cur=cur->next;       }       while(head){           if(head->val!=temp.top())           {               return false;           }           head=head->next;           temp.pop();       }       return true;            }};

我们使用快慢指针找中点的原理是fast和slow两个指针，每次快指针走两步，慢指针走一步，等快指针走完时，慢指针的位置就是中点。我们还需要用栈，每次慢指针走一步，都把值存入栈中，等到达中点时，链表的前半段都存入栈中了，由于栈的后进先出的性质，就可以和后半段链表按照回文对应的顺序比较了。

class Solution {public:    bool isPalindrome(ListNode* head) {        if (!head || !head->next) return true;        ListNode *slow = head, *fast = head;        stack<int> s;        s.push(head->val);        while (fast->next && fast->next->next) {            slow = slow->next;            fast = fast->next->next;            s.push(slow->val);        }        if (!fast->next) s.pop();        while (slow->next) {            slow = slow->next;            int tmp = s.top(); s.pop();            if (tmp != slow->val) return false;        }        return true;    }};

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