LeetCode——Reverse Integer

Reverse Integer

public class Solution {
public int reverse(int x) {
String tmp = new String();
String tmp_2 = new String();
tmp = ((Integer) x).toString();

if (tmp.charAt(0) == '-') {
for (int i = tmp.length() - 1; i > 0; i--)
tmp_2 = tmp_2 + tmp.charAt(i);
tmp = "-" + tmp_2;
} else {
for (int i = tmp.length() - 1; i >= 0; i--)
tmp_2 = tmp_2 + tmp.charAt(i);
tmp = tmp_2;
}

return Integer.parseInt(tmp);
}
}

返回结果超出int整数范围，应该考虑转化之后数值大小的边界检查

原先的思路是用穷举法对结果检查，试了很多次都没有pass；一怒之下，直接用try/catch捕捉异常，搞定。

Java代码：

public class Solution {
public int reverse(int x) {
String tmp = new String();
String tmp_2 = new String();
tmp = ((Integer) x).toString();

if (tmp.charAt(0) == '-') {
for (int i = tmp.length() - 1; i > 0; i--)
tmp_2 = tmp_2 + tmp.charAt(i);
tmp = "-" + tmp_2;
} else {
for (int i = tmp.length() - 1; i >= 0; i--)
tmp_2 = tmp_2 + tmp.charAt(i);
tmp = tmp_2;
}
try{
Integer.parseInt(tmp);
}catch(Exception e){
return 0;
}
return Integer.parseInt(tmp);
}
}