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false if (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long long 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。
