1、題目

Reverse digits of an integer.

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

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Note:
The input is assumed to be a 32-bit signed integer. Your function should?return 0 when the reversed integer overflows.

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2、代碼實現

代碼實現1、 通過不了LeetCode public static int reverse(int x) {if (x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) {return 0;}boolean flag = false; //x是負數就是true，正數falseif (x < 0) {flag = true;}String string = String.valueOf(x);String spliteStr = flag ? string.substring(1) : string;StringBuffer sb = new StringBuffer(spliteStr);sb = sb.reverse();String result = sb.toString();result = flag ? "-" + result : result;long value = Long.valueOf(result);if (value > Integer.MAX_VALUE || value < Integer.MIN_VALUE) {return 0;}return (int)value;}
代碼實現二、 通過不了LeetCode public static int reverse1(int x) {if (x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) {return 0;}long result = 0;int temp = Math.abs(x);while (temp > 0) { result *= 10;result = result + temp % 10;temp /= 10;}if (result > Integer.MAX_VALUE ||result<Integer.MIN_VALUE ) {return 0;}return (int)(x >= 0 ? result : -result);}
? 代碼實現三 可以通過LeetCode public static int reverse3(int n) {if (n > Integer.MAX_VALUE || n < Integer.MIN_VALUE) {return 0;}//輸出結果定義為longlong sum=0;while (n != 0) {int s = n % 10;sum = sum * 10 + s;n = n / 10;}//防止溢出操作if (sum > Integer.MAX_VALUE || sum < Integer.MIN_VALUE) {return 0;}return (int)sum;}
注意有溢出問題，對比分析，第一個實現和第二個實現 不越界沒問題。

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