leetcode 第九题 Palindrome Number(java) Palindrome Number time=434ms 负数不是回文数 public class Solution { public boolean isPalindrome(int x) { int palindrome=0; int revers=x; if(revers<0) return false; else{ while(revers>0){ int m=revers%10; palindrome=m+palindrome*10; revers=revers/...
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome. 说明:从左向右为-121,从右向左为121-,所以它不是回文数 Example 3: Input: 10 Output: false Explanation: Reads 01 from right to left. Therefore it is not a palindrome...
Therefore it is not a palindrome.说明:从右向左读为01,因此它不是回文数Follow up: Coud you solve it without converting the integer to a string你能不能将整数换为字符串?代码class Solution { public boolean isPalindrome(int x) { if(x<0) {return false; } if(x==0) {return true; } Strin...
一、题目 Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. Example 1: Input: 121 Output: true Example 2: Input: -121 Output: false Explanation: From left to right, it reads -121. From right to left, it becomes 121-....
classSolution {publicbooleanisPalindrome(intx) {if(x<0//negative number|| (x%10 == 0 && x != 0))//the check"x == reverseNum/10" has one exception: reverseNum is one-bit number but x isn't, this case only exist in x%10 == 0 && x != 0returnfalse;intreverseNum = 0;whil...
package javaLeetCode_primary; import java.util.Scanner; import java.util.Stack; public class PalindromeNumber_9 { public static void main(String[] args) { @SuppressWarnings("resource") Scanner input = new Scanner(System.in); System.out.println("Please input the integer:"); ...
I have this method, isPalindrome(), and I am trying to find the time complexity of it, and also rewrite the code more efficiently. boolean isPalindrome(String s) { boolean bP = true; for(int i=0; i<s.length(); i++) { if(s.charAt(i) != s.charAt(s.length()-i-1)) { bP...
* In Java How to Check if Number/String is Palindrome or not? */ public class CrunchifyFindPalindromeNumber { public static void main(String[] args) { crunchifyFindPalindromeUsingForLoop(24642); crunchifyFindPalindromeUsingWhileLoop(34567); crunchifyFindPalindromeUsingForLoop(987656789); crunchify...
findRepeatNumber findRepeatNumber2 gameOfNumber gcd&lcm getIntersectionNode getKthFromEnd hasCycle hexadecimalToDecimal hexadecimalToOctal hibernateUtils inorderTraversal isPalindrome isUnique isValid josephRing knapsack kthToLast leapYearJudgment levelOrder ...
所以这里性能也不是很好 // 如果对性能要求比较高的话,还是通过循环从两侧向中间逐一比较,会更好一点 题目中还有一个要求:删除一个字符...,也就是允许一个字符的不同。...code: 出现一处不同 将值传入一个新函数,再进行判断字符串: const validPalindrome = s => { let left = 0; let right = s....