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leetcode : reverse integer
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer‘s last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Update (2014-11-10):
Test cases had been added to test the overflow behavior.
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byte的取值范围为-128~127，占用1个字节（-2的7次方到2的7次方-1）
short的取值范围为-32768~32767，占用2个字节（-2的15次方到2的15次方-1）
int的取值范围为（-2147483648~2147483647），占用4个字节（-2的31次方到2的31次方-1）
long的取值范围为（-9223372036854774808~9223372036854774807），占用8个字节（-2的63次方到2的63次方-1）
转：
（1） 取模的方式，逐个从右往左reverse。
（2）防止溢出，用float先保存。
public class Solution {
public int reverse(int x) {
long rev = 0;
while(x != 0) {
rev = rev * 10 + x % 10;
x = x / 10;
if(rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE) {
return 0;
}
}
return (int) rev;
}
}
leetcode : reverse integer
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