题目描述:
请判断一个链表是否为回文链表。
示例 1:
输入: 1->2
输出: false
示例 2:
输入: 1->2->2->1
输出: true
进阶:
你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题?
解法:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverse(ListNode* head){
if(!head || !head->next){
return head;
}else{
ListNode* cur = head, * nxt = head->next;
head->next = NULL;
while(nxt){
cur = nxt;
nxt = nxt->next;
cur->next = head;
head = cur;
}
return head;
}
}
bool isPalindrome(ListNode* head) {
if(!head || !head->next){
return true;
}else{
int sz = 0;
ListNode* cur = head;
while(cur){
sz++;
cur = cur->next;
}
int half = (sz + 1)/2;
cur = head;
for(int i = 1; i < half; i++){
cur = cur->next;
}
ListNode* rh = cur->next;
cur->next = NULL;
rh = reverse(rh);
cur = head;
while(cur && rh && cur->val == rh->val){
cur = cur->next;
rh = rh->next;
}
return rh == NULL;
}
}
};