leetcode isPalindrome (回文数判断) 回文很简单,就是正着读和反着读一样,要判断一个数是否为回文数只需要判断正反两个是不是相等即可。 再往深了想一下,只需要判断从中间分开的两个数一个正读,一个反读相等即可。 代码: 1 2 3 4 5 6 7 8 9 10 11 12 classSolution { publicbooleanisPalindrome(int...
LeetCode:Valid Palindrome Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. For example, "A man, a plan, a canal: Panama" is a palindrome. "...(leetcode)Valid Palindrome Question Given a string, determine if it is a ...
1classSolution(object):2defisPalindrome(self, x):3"""4:type x: int5:rtype: bool6"""7x2 = str(x)8ifx2 == x2[::-1]:9returnTrue10else:11returnFalse 一个比较精简的代码 运行时间打败了97%的代码 但是很占内存
classSolution{publicbooleanisPalindrome(ListNode head){ListNodeslow= head, fast = head, pre =null;while(fast !=null&& fast.next !=null){ListNodetemp= slow.next;if(pre !=null) slow.next = pre; pre = slow; slow = temp; fast = fast.next.next;}if(fast !=null) slow = slow....
#链接:https://leetcode-cn.com/problems/palindrome-number/solution/jing-xin-hui-zong-python3de-5chong-shi-xian-fang-f/classSolution:# 方法一:将int转化成str类型:双向队列 # 复杂度:O(n^2)[每次pop(0)都是O(n)..比较费时]defisPalindrome(x:int)->bool:lst=list(str(x))whilelen(lst)>1:...
© 2025 领扣网络(上海)有限公司 3K 5.4K 0 人在线 1 2 3 4 5 6 classSolution{ public: boolisPalindrome(intx) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2Case 3 x = 121 9 1 2 3 › 121 -121 10 Source
[LeetCode] Palindrome Partitioning II, Solution Given a strings, partitionssuch that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning ofs. For example, givens="aab", Return1since the palindrome partitioning["aa","b"]could be produced...
class Solution { public boolean isPalindrome(String s) { StringBuffer sgood = new StringBuffer(); int length = s.length(); for (int i = 0; i < length; i++) { char ch = s.charAt(i); if (Character.isLetterOrDigit(ch)) { sgood.append(Character.toLowerCase(ch)); } } String...
class Solution {public: bool isPalindrome(string s) { if (s.empty()) return true; const char* s1 = s.c_str(); const char* e = s1 + s.length() - 1; while (e > s1) { if (!isalnum(*s1)) {++s1; continue;} if (!isalnum(*e)) {--e; continue;} ...
Have you consider that the string might be empty? This is a good question to ask during an interview. For the purpose of this problem, we define empty string as valid palindrome. SOLUTION 1: 左右指针往中间判断。注意函数: toLowerCase ...