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.
（1）思想1：设立一个falg=1，判断x<0，则令flag=-1；然后仍然是将数字转换成字符串，进行str[i] 与 str[len-1-i] 的一个互换，互换完成之后再将字符串转化成数字，此时需要注意的是：数字的范围是-2^32 ~ 2^32-1，所以必须对转换前和转换后的数字均进行判断，在这个范围内，则返回 flag*转换后的数字，否则返回0。
C++：
1 class Solution { 2 public: 3 int reverse(int x) { 4 int flag=1; 5 if(x<0) 6 flag=-1; 7 string str=to_string(abs(x)); 8 int len=str.size(); 9 for(int i=0;i<len/2;i++) 10 { 11 char c=str[i]; 12 str[i]=str[len-1-i]; 13 str[len-1-i]=c; 14 } 15 long int result = flag*atol(str.c_str()); 16 if(x>pow(2,31)-1 || x<-1*pow(2,31) || result>pow(2,31)-1 || result<-1*pow(2,31)) 17 return 0; 18 return result; 19 } 20 };