【LeetCode】7. Reverse Integer

最新推荐文章于 2022-06-06 18:52:41 发布

weixin_34023982

最新推荐文章于 2022-06-06 18:52:41 发布

阅读量76

点赞数

本文介绍了一种用于反转整数的算法实现，通过将输入的整数转换为64位数来避免溢出问题，并逐步构建反转后的整数，同时检查结果是否超出32位整数的范围。

Reverse Integer

Reverse digits of an integer.

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

click to show spoilers.

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).

（1）首先将x提升为64位num，若不然-INT_MIN将会溢出.

（2）将num取模10得到的每一位数字加入到返回值中，然后返回值乘10移位，得到数值上的结果ret

（3）判断ret是否超过int的范围，若超过，则返回0

时间复杂度为：O(k), k为x的位数

class Solution {
public:
    int reverse(int x) {
        long long ret = 0;
        int sign = (x<0)?-1:1;
        x = abs(x);
        while(x)
        {
            int digit = x % 10;
            x /= 10;
            ret = ret * 10 + digit;
        }
        ret *= sign;
        if(ret > INT_MAX || ret < INT_MIN)
            return 0;
        else
            return ret;
    }
};