LeetCode 回文链表

题目链接：https://leetcode-cn.com/problems/palindrome-linked-list/

题目大意

　　略。

分析

　　反转后半条链表，再判断可实现空间复杂度 O(1)。

代码如下

/**
 * 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) {
        ListNode *p1, *p2;
        int len = getLen(head);
        
        p1 = head;
        for(int i = 1; i <= (len - 1) >> 1; ++i) p1 = p1->next;
        p2 = reverseList(p1);
        p1 = head;
        
        while(p1 != NULL && p2 != NULL) {
            if(p1->val != p2->val) return false;
            p1 = p1->next;
            p2 = p2->next;
        }
        return true;
    }
    
    ListNode* reverseList(ListNode* head) {
        if(head == NULL || head->next == NULL) return head;
        
        ListNode *p1, *p2;
        p1 = head;
        p2 = head->next;
        p1->next = NULL;
        
        while(p2 != NULL) {
            ListNode *tmp = p2->next;
            
            p2->next = p1;
            p1 = p2;
            p2 = tmp;
        }
        
        return p1;
    }
    
    int getLen(ListNode* x) {
        int ret = 0;
        while(x != NULL) {
            ++ret;
            x = x->next;
        }
        return ret;
    }
};