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?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

分析

题目要求将一个int逆序，如123逆序为321，-123逆序为-321.

1、如果输入的是负数，那么逆序的int也是负的

2、逆序的数字首位有0，需要忽略。如009是应该写作9的。

     * 虽然题目note里面说了要注意整形溢出，但是给出的函数类型返回值是int，code就不用考虑溢出了~
     * 输入整形的正负号，我们用一个变量@neg来标记。
     * 在字符串和整形之间的转化中，我们统一都将输入整数当作正数来看待（负的取绝对值）
     * 借助于stringstream，我们将int转为string，再把string逆序，最后再借助于stringstream，
     * 将string转为int（不必自己去手动忽略string的低位0）。

Code

class Solution {
public:
    /* 
     * 虽然题目note里面说了要注意整形溢出，但是给出的函数类型返回值是int，code就不用考虑溢出了~
     * 输入整形的正负号，我们用一个变量@neg来标记。
     * 在字符串和整形之间的转化中，我们统一都将输入整数当作正数来看待（负的取绝对值）
     * 借助于stringstream，我们将int转为string，再把string逆序，最后再借助于stringstream，
     * 将string转为int（不必自己去手动忽略string的低位0）。
     * */
    int reverse(int x) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int neg = 1;
        if(x < 0)
            neg = -1, x = -x;
        stringstream ss;
        string s;
        char temp;
        int integer;
        ss<<x;
        ss>>s;
        for(int pos = 0; pos < s.size()/2; pos++)
        {
            temp = s[pos];
            s[pos] = s[s.size() - 1 -pos];
            s[s.size() - 1 -pos] = temp;
        }
        // 再次使用stringstream，要先clear
        ss.clear();
        ss<<s;
        ss>>integer;
        return neg*integer;
    }
};