1、题目
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、代码实现
代码实现1、
通过不了LeetCode
public static int reverse(int x) {if (x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) {return 0;}boolean flag = false; //x是负数就是true，正数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;}代码实现二、
通过不了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);}代码实现三
可以通过LeetCode
public static int reverse3(int n) {if (n > Integer.MAX_VALUE || n < Integer.MIN_VALUE) {return 0;}//输出结果定义为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;}注意有溢出问题，对比分析，第一个实现和第二个实现 不越界没问题。
