題目描述
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
click to show spoilers.
Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

解析：看了題目描述了這么多廢話，無非是輸入10返回1，輸入-21返回-12等等。解析思路就是該怎么來就怎么來，翻轉數字，注意負號的判斷就好了。

public class Solution {public int reverse(int x) {if(x == 0){return x;}while (x%10==0){x/=10;}String s=String.valueOf(x);boolean flag=false;StringBuilder sb = new StringBuilder();for(int i=s.length()-1;i>=0;i--){if(i==0){if(s.charAt(i)=='-'){flag=true;continue;}}sb.append(s.charAt(i));}return flag==true?-1*Integer.parseInt(sb.toString()):Integer.parseInt(sb.toString());} }

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