No.009 Palindrome Number

时间:2022-04-29
本文章向大家介绍No.009 Palindrome Number,主要内容包括其使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

9. Palindrome Number

  • Total Accepted: 136330
  • Total Submissions: 418995
  • Difficulty: Easy

  Determine whether an integer is a palindrome. Do this without extra space.

  Some hints:

  Could negative integers be palindromes? (ie, -1)

  If you are thinking of converting the integer to string, note the restriction of using extra space.

  You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?

  There is a more generic way of solving this problem.

  思路:

  本题解题很简单,首先判断负数和0,然后我们计算得到最大基数,如果x为一个两位数,则基数base=100,三位数则为1000等,然后我们每次分别取这个数的最高位和最低位进行比较,如果不相等,则直接返回false,相等则去掉最左边和最右边的数后进行下一轮比较。代码如下:

 1 public boolean isPalindrome(int x) {
 2     if(x < 0){
 3         return false ;
 4     }
 5     if(x == 0 ){
 6         return true ;
 7     }
 8     
 9     int base = 1 ;
10     while(x/base >= 10){
11         base *= 10 ;
12     }
13     
14     while(x != 0){
15         int leftDigit = x/base ;
16         int rightDigit = x%10 ;
17         if(leftDigit != rightDigit){
18             return false ;
19         }
20         x = x%base/10 ;
21         base /= 100 ;
22     }        
23     return true ;      
24 }