[leetcode] Palindrome Linked List

最新推荐文章于 2022-04-04 23:52:01 发布

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本文介绍了一种O(n)时间复杂度和O(1)空间复杂度的算法来判断单链表是否为回文串。通过反转链表的一半并比较两部分元素的方式实现。

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?

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
    	ListNode[] ll = reverseHalf(head);
    	ListNode l1 = ll[0];
    	ListNode l2 = ll[1];
    	while(null != l1) {
    		if(l1.val != l2.val) {
    			return false;
    		}
    		l1 = l1.next;
    		l2 = l2.next;
    	}
    	return true;
    }

	private ListNode[] reverseHalf(ListNode head) {
		ListNode[] ll = {null, null};
		ListNode fast=head, slow=head, pre=null;
		ListNode ph = new ListNode(0);
		while(null != fast && null != fast.next) {
			pre = slow;
			slow = slow.next;
			fast = fast.next.next;
			pre.next = ph.next;
			ph.next = pre;
		}
		if(null != pre) {
			ll[0] = ph.next;;
			if(null != fast) {
				ll[1] = slow.next;;
			} else {
				ll[1] = slow;			
			}
		}
 		return ll;
	}
}

/**
 * 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 *l1=NULL, *l2=NULL;
        reverseHalf(head, l1, l2);
    	while(l1) {
    		if(l1->val != l2->val) {
    			return false;
    		}
    		l1 = l1->next;
    		l2 = l2->next;
    	}
    	return true;
    }
	void reverseHalf(ListNode* head, ListNode* &l1, ListNode* &l2) {
		ListNode *fast=head, *slow=head, *pre=NULL;
		ListNode *ph = new ListNode(0);
		while(fast && fast->next) {
			pre = slow;
			slow = slow->next;
			fast = fast->next->next;
			pre->next = ph->next;
			ph->next = pre;
		}
		if(pre) {
			l1 = ph->next;;
			if(fast) {
				l2 = slow->next;;
			} else {
				l2 = slow;			
			}
		}
		delete ph;
	}
};