[Leetcode] #234 Palindrome Linked List

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Discription:

Given a singly linked list, determine if it is a palindrome.Could you do it in O(n) time and O(1) space?

Solution:

将链表值记录成数组，判断数组是否回文。时间复杂度O(n)，空间复杂度O(n)。

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

将链表后半段反转与前半段比较，给出两种写法。用双指针找链表中点，注意链表结点是奇数个还是偶数个。时间复杂度O(n)，空间复杂度O(1)。

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

class Solution {public:    ListNode *reverseList(ListNode *head){        ListNode *pre = NULL, *temp;        while(head){            temp = head->next;            head->next = pre;            pre = head;            head = temp;        }        return pre;    }    bool isPalindrome(ListNode* head) {        if(!head || !head->next) return true;        ListNode *slow = head,*fast = head;        while(fast && fast->next){            fast = fast->next->next;            slow = slow->next;        }        if(fast){//链表个数为奇数个            slow = reverseList(slow->next);        }        else{//链表个数为偶数个            slow = reverseList(slow);        }        while(slow){            if(head->val != slow->val)                return false;            head = head->next;            slow = slow->next;        }        return true;    }};

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