LeetCode7-Reverse Integer(C++)

Description

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 store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

AC代码

class Solution {
public:
    int reverse(int x) {
        long long ret=0;
        while(x != 0) {
            ret = ret * 10 + x % 10;
            x /= 10;
            if((ret > INT32_MAX) || (ret < INT32_MIN)) {
                return 0;
            }
        }
        return ret;
    }
};

测试代码

int main() {
    Solution s;
    cout << s.reverse(1534236469) << " " << s.reverse(-123) << endl;
}

总结

反转整数是一道比较经典的算法题目，处理思路就是通过%10一步步把整数的最后一位剥离出来，剩下的数就是/10。再对剩下的数进行相同的操作，直到x=0为止。