#413 Reverse Integer

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题目描述：

Reverse digits of an integer. Returns 0 when the reversed integer overflows (signed 32-bit integer).

Have you met this question in a real interview?

Yes

Example

Given x = 123, return 321

Given x = -123, return -321

题目思路：

这题主要注意的是sign和overflow的问题，没什么trick，直接上code了。

Mycode（AC = 30ms）：

class Solution {public:    /**     * @param n the integer to be reversed     * @return the reversed integer     */    int reverseInteger(int n) {        // Write your code here        int sign = n >= 0? 1 : -1;        long long long_n = abs((long long)n);                // get the reversed integer in string        string str_n = to_string(long_n);        reverse(str_n.begin(), str_n.end());                // compute the reversed integer in long long        long_n = 0;        for (int i = 0; i < str_n.length(); i++) {            long_n = long_n * 10 + int(str_n[i]) - int('0');            if (abs(long_n) >= abs((long long)INT_MIN)) {                return 0;            }        }                // return values        long_n *= sign;        if (long_n > INT_MAX || long_n < INT_MIN) {            return 0;        }        else {            return (int)long_n;        }    }};

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