题目 c++ class Solution { public: int a[100005]; bool isPalindrome(int x) { if(x<0) return false; int pos=0; while(x) { a[pos++]=x%10; x/=10; }
class Solution { public: bool isPalindrome(int x) { //negative number if(x < 0) len >= 10) len *= 10; while(x > 0) { //get the head and tail number right) return false; else { //remove the head and tail number
Determine whether an integer is a palindrome. Do this without extra space. click to show spoilers.
Problem # Determine whether an integer is a palindrome. Do this without extra space. # # click to sh
Palindrome Number Determine whether an integer is a palindrome.
:type x: int 5 :rtype: bool 6 """ 7 temp = x 8 revert_x = 0 9
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/palindrome-number 著作权归领扣网络所有。
Determine whether an integer is a palindrome. Do this without extra space. 解题思路: 转化为字符串,然后反转判断与原字符串是否相同 Python实现: class Solution(object): def isPalindrome(self, x): """ :type x: int :rtype: bool """ if x < 0: #
第一种方法:网友的思路,翻转一半能够避免21亿零9这样的数字溢出问题,很巧妙,但是9000000012这个数字溢出后也和原来的数字不等,int型的溢出在这里并不会造成影响 class Solution
are not palindrome if (x < 0 || x % 10 == 0) return false; // reverse half of the number
这道题目很简单,只有一句话,不要要求不使用额外空间,一般来说不使用额外空间的意思是不使用复杂度为O(n)的额外空间,新建一些字符串、整型值之类的还是可以的。回文的意思是从左到右读和从右到左读数字是一样的,比如11是回文,121是回文。
最常规的思路是直接将数反转再对比,但实际上,我们只需要反转后半部分跟前半部分对比就可以了。两种情况,一是位数为偶数,直接对比,位数为奇数,大数除以10对比。
提交中击败了19.58%的用户 leetcode执行: Runtime: 28 ms, faster than 21.90% of Go online submissions for Palindrome Number Memory Usage: 5.6 MB, less than 14.02% of Go online submissions for Palindrome Number.
版权声明:原创勿转 https://blog.csdn.net/anakinsun/article/details/88867332
题目地址:https://leetcode.com/problems/palindrome-number/description/ 题目要求: Determine whether an integer
关关的刷题日记79 – Leetcode 9 Palindrome Number 题目 Determine whether an integer is a palindrome.
6 int i , xor; 7 for(xor = 0 , i = 0 ; i < length ; ++i) 8 xor = xor ^ arr[i]; 9
题目 Given a non-negative integer num, return the number of steps to reduce it to zero. If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.
现在平台有个字段是用来记录插入时间的,但是是用number型存储,想转为时间类型的。 解决问题的过程: http://blog.csdn.net/a9529lty/article/details/5306622 ORACLE 毫秒转换为日期 日期转换毫秒 日期转换毫秒 SELECT TO_NUMBER TO_DATE(‘1970-01-01 8:0:0’, ‘YYYY-MM-DD HH24:MI:SS’)) * 24 * 60 * 60 * 1000 FROM DUAL; SELECT TO_NUMBER
7 eval(str ) 用来计算在字符串中的有效Python表达式,并返回一个对象 8 tuple(s ) 将序列 s 转换为一个元组 9