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.
Tags: Math
题意是给你一个整型数，求它的逆序整型数，而且有个小坑点，当它的逆序整型数溢出的话，那么就返回 0，用我们代码表示的话可以求得结果保存在 long 中，最后把结果和整型的两个范围比较即可。
class Solution {
public int reverse(int x) {
long res = 0;
for (; x != 0; x /= 10)
res = res * 10 + x % 10;
return res > Integer.MAX_VALUE || res < Integer.MIN_VALUE ? 0 : (int) res;
}
}如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：awesome-java-leetcode