LeetCode234给你一个单链表的头节点head请你判断该链表是否为回文链表。如果是返回true否则返回false。回文序列是向前和向后读都相同的序列。Python解法1.变成数组# Definition for singly-linked list. # class ListNode: # def __init__(self, val0, nextNone): # self.val val # self.next next class Solution: def isPalindrome(self, head: Optional[ListNode]) - bool: res [] current head while current is not None: res.append(current.val) current current.next return res res[::-1]2.快慢指针# Definition for singly-linked list. # class ListNode: # def __init__(self, val0, nextNone): # self.val val # self.next next class Solution: def get_mid(self, head): slow head fast head while fast.next is not None and fast.next.next is not None: slow slow.next fast fast.next.next return slow def reverse_list(self,node): cur node pre None while cur: nxt cur.next cur.next pre pre cur cur nxt return pre def isPalindrome(self, head: Optional[ListNode]) - bool: if not head or not head: return True mid_node self.get_mid(head) second_half mid_node.next reverse_second self.reverse_list(second_half) p1 head p2 reverse_second flag True while flag and p2: if p1.val ! p2.val: flag False p1 p1.next p2 p2.next return flagJava解法1.变成数组/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val val; } * ListNode(int val, ListNode next) { this.val val; this.next next; } * } */ class Solution { public boolean isPalindrome(ListNode head) { int len 0; ListNode cur head; while(cur ! null){ len; cur cur.next; } int[] arr new int[len]; cur head; for(int i 0; i len; i){ arr[i] cur.val; cur cur.next; } int s 0, f len - 1; while(s f){ if(arr[s] ! arr[f])return false; s; f--; } return true; } }2.快慢指针/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val val; } * ListNode(int val, ListNode next) { this.val val; this.next next; } * } */ class Solution { // 快慢指针找前半段中点 private ListNode getMid(ListNode head) { ListNode slow head; ListNode fast head; while (fast.next ! null fast.next.next ! null) { slow slow.next; fast fast.next.next; } return slow; } // 反转链表 private ListNode reverseList(ListNode node) { ListNode cur node; ListNode pre null; while (cur ! null) { ListNode nxt cur.next; cur.next pre; pre cur; cur nxt; } return pre; } public boolean isPalindrome(ListNode head) { if (head null || head.next null) { return true; } ListNode midNode getMid(head); ListNode secondHalf midNode.next; ListNode revSecond reverseList(secondHalf); ListNode p1 head; ListNode p2 revSecond; boolean flag true; while (flag p2 ! null) { if (p1.val ! p2.val) { flag false; } p1 p1.next; p2 p2.next; } return flag; } }C解法1.变成数组/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */ class Solution { public: bool isPalindrome(ListNode* head) { vectorint res; ListNode* cur head; while(cur ! nullptr){ res.push_back(cur-val); cur cur-next; } vectorint rev(res.rbegin(), res.rend()); return res rev; } };2.快慢指针/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */ class Solution { private: // 找中点 ListNode* getMid(ListNode* head) { ListNode* slow head; ListNode* fast head; while (fast-next ! nullptr fast-next-next ! nullptr) { slow slow-next; fast fast-next-next; } return slow; } // 反转链表 ListNode* reverseList(ListNode* node) { ListNode* cur node; ListNode* pre nullptr; while (cur ! nullptr) { ListNode* nxt cur-next; cur-next pre; pre cur; cur nxt; } return pre; } public: bool isPalindrome(ListNode* head) { if (head nullptr || head-next nullptr) { return true; } ListNode* midNode getMid(head); ListNode* secondHalf midNode-next; ListNode* revSecond reverseList(secondHalf); ListNode* p1 head; ListNode* p2 revSecond; bool flag true; while (flag p2 ! nullptr) { if (p1-val ! p2-val) { flag false; } p1 p1-next; p2 p2-next; } return flag; } };