Leetcode ☞ 234. Palindrome Linked List

234. Palindrome Linked List

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Question

Total Accepted: 40299  Total Submissions: 147797  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?

我的AC：（14ms，最快一批，击败25%）

bool isPalindrome(struct ListNode* head) {
    if (!head || !head->next) return true;//不加就runtime error
    
    struct ListNode *fast, *slow, *mid;
    fast = slow = head;
    while(fast->next && fast->next->next){
        fast = fast->next->next;
        slow = slow->next;
    }
    mid = slow->next;
    //[1,2,3,4]结束时，fast为3，slow为2【mid为3，正确】；[1,2,3,4,5]结束时，fast为5，slow为3【mid为4，正确】。
    //所以循环的条件一个是fast->next，另一个是fast->next->next而非fast。
    //不然，[1,2,3,4]结束时，fast为NULL，slow为3【mid为4，错误】；[1,2,3,4,5]结束时，fast为5，slow为3【mid为4，正确】。
    struct ListNode *p, *p1, *p2;
    p2 = mid;
    p1 = NULL;
    while(p2){
        p = p2->next;
        p2->next = p1;
        p1 = p2;
        p2 = p;
    }
    mid = p1;//此时原链表的后半部反转完成，接下来对前/后半部逐个对比。
    while(mid){
        if(mid->val != head->val)
            return false;
        mid = mid->next;
        head = head->next;
    }
    return true;
}

思路：

1、快慢指针找到中位数

2、反转后半部分

3、将head跟mid逐个对比

一定别忘if (!head || !head->next) return true;  ！！！！

附找中位数代码（快慢指针）：

struct ListNode *fast, *slow;  
fast = slow = head;  
if (!head || !head->next) return head;  
  
while (fast&&slow)  
{  
　　if (fast->next==NULL)  
　　    return slow ->data;  
　　else if (fast->next!= NULL && fast->next->next== NULL)  
　　    return (slow ->data + slow ->next->data)/2;  
　　else{  
　　    fast= fast->next->next;  
　　    slow = slow ->next;  
　　}  
}