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, Solution 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."race a car" is not a palindrome....
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
class Solution(object): def isPalindrome(self, x): """ :type x: int :rtype: bool """ if x < 0: return False temp = x y = 0 while temp: y = y*10 + temp%10 temp /= 10 return x == y 总结 由于不允许占用额外空间,所以不能将其分为字符串来做。 本文参与 腾讯云自媒体同步曝...
[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...
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 ...
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...
classSolution {public:boolisPalindrome(intx) {inttemp=x,n=0,len=0,i;if(x<0)returnfalse;while(temp>0)//计算数字长度{ temp/=10; len++; }for(i=0;i<len/2;i++)//二分, 高位保留,低位翻转相乘{ n= n*10+ x%10; x/=10;
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....
classSolution {publicbooleanisPalindrome(intx) {if(x<0){returnfalse; }else{inty = 0;intz =x;while(x/10!=0){ y*=10; y+=x%10; x/=10; } y=y*10+x;if(y==z){returntrue; }else{returnfalse; } } } } 负数都不成功,所以直接报false...