LeetCode Reverse Integer

Reverse Integer Total Accepted: 61132 Total Submissions: 219035 My Submissions Question Solution
Reverse digits of an integer.

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

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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?

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

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

题意：反转数字，可是要注意溢出的情况下。还有100的反序，对于溢出的情况下，返回0,100的反序为1而不是001。

先给出第一种解法：
既然是反转，那么最简单的就是把数字看成字符串，然后进行逆序。负号进行特殊处理。然后使用sscanf把字符串化成数字。

对溢出的问题，能够这么想，如果输入是x，如果逆序溢出了的话，x的长度至少是10位。此外。如果逆序后的字符串为s1，sscanf后的数字为y，那么这个y相应的字符串s2将和s1不相等！

代码例如以下：

class Solution {
public:
    int reverse(int x) {
        int result=0;
        char s[12];
        char r[12];
        sprintf(s,"%d",x);
        int i=0;
        if(s[0]=='-'){
            r[0]='-';
            i++;
        }
        r[strlen(s)]='\0';
        int j=strlen(s)-1;
        while(i<strlen(s))
        {
            r[i]=s[j];
            j--;
            i++;
        }
        sscanf(r,"%d",&result);

        if(strlen(s)>=10)//溢出的可能
        {
            sprintf(s,"%d",result);
            if(strcmp(r,s))
                result=0;
        }
        return result;
    }
};

1032 / 1032 test cases passed.
Status: Accepted
Runtime: 16 ms

另外一种解法：
额， 事实上这才是最简单的方法。通过重复将个位上的数字前移就好了，如果溢出直接返回0！

class Solution {  
public:  
    int reverse(int x) {    
        const int max = 0x7fffffff;  //int最大值  
        const int min = 0x80000000;  //int最小值  
        long long sum = 0;   

        while(x != 0)  
        {  
            int temp = x % 10;  
            sum = sum * 10 + temp;  
            if (sum > max || sum < min)   //溢出处理  
                return 0;  
            x = x / 10;  
        }  
        return sum;  
    }  
}; 

1032 / 1032 test cases passed.
Status: Accepted
Runtime: 14 ms