【easy】234. Palindrome Linked List

ques: 判断一个链表是否回文 Could you do it in O(n) time and O(1) space?

method：先将链表分为两部分，将后半部分反转，最后从前往后判断是否相等。

topic: 链表，链表反转

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */  class Solution {
public:bool isPalindrome(ListNode *head) { //input check  abcba abccba  if(head == NULL || (head&&head->next==NULL)) return true;  ListNode *middle = partition(head);  middle = reverse(middle);  while(head!=NULL && middle!=NULL) {  if(head->val != middle->val) return false;  head = head->next;  middle = middle->next;  }  return true;  }
private: ListNode* partition(ListNode *head) {  ListNode* p = head;  while(p&&p->next!=NULL && p->next->next!=NULL) {  p = p->next->next;  head = head->next;  }  p = head->next;  head->next = NULL;  return p;  }
private: ListNode* reverse(ListNode *head) {  if(head==NULL || head->next==NULL) return head;  ListNode* pre = head;  ListNode* cur = head->next;  pre->next = NULL;  ListNode* nxt = NULL;  while(cur!=NULL) {  nxt = cur->next;  cur->next = pre;  pre = cur;  cur = nxt;  }  return pre;  }
};

转载于:https://www.cnblogs.com/sherry-yang/p/8245187.html

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