Reverse Integer--整数的反转

原题：

Reverse digits of an integer.

=>反转一个整数的数字。例子如下：

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

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!

=>在做题的时候请仔细思考一下下面这些方面。

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

=>假如最后一位是0，那么结果会是什么样的呢？比如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?

=>你有考虑过反转后的溢出问题嘛？假如是32bit的整数，1000000003反转后就会溢出。你怎么处理这样的情况？

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).

=>抛出一个异常？很好，假如不能抛出异常呢？其实可以重新定义这个函数（比如加一个参数）

class Solution {
public:
    int reverse(int x) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
 
};

晓东分析：

这个题目就反转本身而言是很简单的，晓东就不多分析了。所以，我主要来说一下溢出的问题，我个人的思路就是在得到反转的值的时候，先不乘上最高位，留着进行比较。

所以总的来说，还是不复杂的。

代码实现：

class Solution {
public:
    int reverse(int x) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int max_first_bit = 2; //default for 4
        int max_remain_num = 147483647;
 
        int num = 0;
        int temp = abs(x);
 
        while(temp >= 10){
            num = num * 10 + temp % 10;
            temp /= 10;
        }
 
        switch(sizeof(int)){
            case 1:
                max_first_bit = 1;
                max_remain_num = 27;
                break;
            case 2:
                max_first_bit = 3;
                max_remain_num = 2767;
                break;
            case 4:
                max_first_bit = 2;
                max_remain_num = 147483647;
                break;
            case 8:
                max_first_bit = 9;
                max_remain_num = 223372036854775807;
                break;
        }    
 
        if(x > 0){
            if (temp < max_first_bit)
                return num * 10 + temp % 10;
            else if(num <= max_remain_num)
                return num * 10 + temp % 10;
            else
                throw x;
        }else{
            if (temp < max_first_bit)
                return 0 - (num * 10 + temp % 10);
            else if(num <= max_remain_num + 1)
                return 0 - (num * 10 + temp % 10);
            else
                throw x;
        }
 
    }
};

执行结果：

1020 / 1020 test cases passed.	Status: Accepted
Runtime: 28 ms

希望大家有更好的算法能够提出来，不甚感谢。

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