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[LeetCode] Reverse Integer

[LeetCode] Reverse Integer

作者: lalulalula | 来源:发表于2017-09-14 11:55 被阅读0次

    1.Reverse digits of an integer.

    Example1: x = 123, return 321
    Example2: x = -123, return -321

    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?

    For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

    2.Note:
    The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

    3.题目要求:翻转数字,需要考虑溢出问题。可以用long long型变量保存计算结果,最后返回的时候判断是否在int返回内。

    4.代码:
    class Solution {
    public:
    int reverse(int x) {
    long long res = 0;
    while (x != 0) {
    res = 10 * res + x % 10;
    x /= 10;
    }
    return (res > INT_MAX || res < INT_MIN) ? 0 : res;
    }
    };

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