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_007_Reverse_Integer.java
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_007_Reverse_Integer.java
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package com.github.junyu.solution.leetCode.easy.string;
/**
* @author ShaoJunyu
* @since 2018/6/14 09:15
*/
public class _007_Reverse_Integer {
/*
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1].
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
*/
/**
* 利用取余和取商得方法,使用while每次对输入得数进行/10操作,直到最后/10得结果为0结束。
* 在循环体中每次对事先定义得result进行*10+余数累增,得出翻转后得结果,加入最大值和最小值
* 得判断防止超出边界
* @param x
* @return
*/
public static int reverse(int x) {
int result = 0;
while (x != 0) {
int remainder = x % 10;
x = x / 10;
if (result > Integer.MAX_VALUE / 10 || result < Integer.MIN_VALUE / 10) {
return 0;
}
result = result * 10 + remainder;
}
return result;
}
public static void main(String[] args) {
int num = 123;
// int num = -123;
// int num = 110;
// int num = -2147483648;
// int num = 2147483647;
System.out.println(reverse(num));
}
}