234 - Palindrome Linked List

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?

沒有達到上述的 O(1) space。

思路：

將 鏈表的前半部分壓入棧中，然後逐個彈出和鏈表的後半部分比較，從而獲知是否迴文。

寫的過程中，主要遇到的問題集中在鏈表長度爲奇偶數的不同，奇數要先彈出棧頂元素。

C++ 代碼如下：

#include <iostream>
#include <stack>
using namespace std;

struct ListNode {
        int val;
        ListNode *next;
        ListNode(int x): val(x), next(NULL) {}
};

class Solution {

public:
        bool isPalindrome(ListNode* head) {
                if(head == NULL || head->next == NULL)
                        return true;

                ListNode * current = head;
                int i, length;
                for( i = 0; current != NULL; i++ ) {
                        current = current->next;
                }
                length = i;  

                stack<int> mystack;
                current = head;
                int limit = length /2 ;
                if(length % 2 != 0)
                        limit ++;
                for(int i = 0; i < limit; i++, current = current->next)
                        mystack.push(current->val);

                if(length % 2 != 0) 
                        mystack.pop();

                for( --i; i < length && current != NULL; current = current->next) {
                        int v;
                        if(!mystack.empty()) {
                                v = mystack.top();
                                mystack.pop();
                        }
                        if(v != current->val) 
                                return false;

                }
                return true;
        }

        ListNode * init() {
                ListNode * head = new ListNode(1);
                ListNode * second = new ListNode(2);
                ListNode * third = new ListNode(2);
                ListNode * forth = new ListNode(1);
                ListNode * fifth = new ListNode(3);
                head->next = second;
                head->next->next = third;
                head->next->next->next = forth;
                head->next->next->next->next = fifth;

                if( isPalindrome(head) == true )
                        cout << "yes" << endl;
                else    cout << "no" << endl;

                while( head != NULL) {
                        cout << head->val << endl;
                        head = head->next;
                }
                return head;
        }
};

int main() {
        Solution s;
        s.init();
        return 0;
}