1、題目：

Reverse digits of an integer.

Example1:?x = 123, return 321
Example2:?x = -123, return -321

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).

2、解答：

public class Solution {public int reverse(int x) {int out = 0;//保存當前結果int newout = 0;//保存最新結果while(x!=0){newout = out*10 + x%10;//目前最新計算的結果=上次結果*10+新的尾數if((newout-x%10)/10!=out){//逆運算判斷是否溢出return 0;}x = x/10;//降位out = newout;}return out;} }

3、思路：

最簡單的做法是：

public class Solution {public int reverse(int x) {int out = 0;while(x!=0){out = out*10 + x%10;x = x/10;}return out;} }

但存在溢出問題，int型有自己的取值范圍。

2的解法加了一步逆運算，

if((newout-x%10)/10!=out)，即用當前的最新結果newout-剛才加上的尾數x%10，若不等于out*10，則說明存在溢出問題。

溢出是指，newout = out*10 + x%10的值超過最大位數時，系統會將超過最大位數的高位舍棄，從而導致反向驗算時出現(newout-x%10)/10!=out。

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