1、题目
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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。
