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 hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
数字颠倒位置，这个比较简单
先来一个奇葩的做法，先把整数变成字符串，然后颠倒一把，然后转变成一个long型的整数，假如出现越界直接返回0，没越界的话，强制转型返回int即可
七拼八凑最终还是Ok的，搞笑方法如下
package com.maoxiaomeng;
/**
* @author lihui
* @date 2018/3/25 22:31
*/
public class Solution {
public int reverse(int x) {
if (x == 0) {
return x;
}
while (x % 10 == 0 && x != 0) {
x = x / 10;
}
String s = x + "";
if (s.charAt(0) == '-') {
s = '-' + new StringBuffer(s.substring(1, s.length())).reverse().toString();
} else {
s = new StringBuffer(s).reverse().toString();
}
long temp = Long.valueOf(s);
if (temp > Integer.MAX_VALUE || temp < Integer.MIN_VALUE) {
temp = 0;
}
return (int)temp;
}
}
下面是正规方法
比如1234 = ((1 * 10 + 2) * 10 + 3) * 10 + 4
那么倒序就是4321 = ((4 * 10 + 3) * 10 + 2) * 10 + 1
得到temp = temp * 10 + x % 10，然后x = x / 10
package com.maoxiaomeng;
/**
* @author lihui
* @date 2018/3/25 22:31
*/
public class Solution {
public int reverse(int x) {
long temp = 0;
while (x != 0) {
temp = temp * 10 + x % 10;
x = x / 10;
}
if (temp > Integer.MAX_VALUE || temp < Integer.MIN_VALUE) {
temp = 0;
}
return (int)temp;
}
}
简单多了
