Reverse Integer--整数的反转
原题：
Reverse digits of an integer.
=>反转一个整数的数字。例子如下：
Example1: x = 123, return 321
Example2: x = -123, return -321
class Solution {public: int reverse(int x) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. };
晓东分析：
这个题目就反转本身而言是很简单的，晓东就不多分析了。所以，我主要来说一下溢出的问题，我个人的思路就是在得到反转的值的时候，先不乘上最高位，留着进行比较。
所以总的来说，还是不复杂的。
代码实现：
class Solution {public: int reverse(int x) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. int max_first_bit = 2; //default for 4 int max_remain_num = 147483647; int num = 0; int temp = abs(x); while(temp >= 10){ num = num * 10 + temp % 10; temp /= 10; } switch(sizeof(int)){ case 1: max_first_bit = 1; max_remain_num = 27; break; case 2: max_first_bit = 3; max_remain_num = 2767; break; case 4: max_first_bit = 2; max_remain_num = 147483647; break; case 8: max_first_bit = 9; max_remain_num = 223372036854775807; break; } if(x > 0){ if (temp < max_first_bit) return num * 10 + temp % 10; else if(num <= max_remain_num) return num * 10 + temp % 10; else throw x; }else{ if (temp < max_first_bit) return 0 - (num * 10 + temp % 10); else if(num <= max_remain_num + 1) return 0 - (num * 10 + temp % 10); else throw x; } }};
执行结果：
1020 / 1020 test cases passed.Status:
希望大家有更好的算法能够提出来，不甚感谢。
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