LeetCode Reverse Integer-优快云博客

本文链接：https://blog.youkuaiyun.com/wdkirchhoff/article/details/44497209

本文介绍了一种用于反转整数的有效算法，并提供了两种实现方法。一种是通过字符串转换和逆序，另一种则是通过数学操作逐位反转。同时讨论了处理溢出情况及特殊情况如100的反转。

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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

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?

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，但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，可以这么想，假设输入是x，如果逆序溢出了的话，x的长度至少是10位，此外，假设逆序后的字符串为s1，sscanf后的数字为y，那么这个y对应的字符串s2将和s1不相等！

代码如下：

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

第二种解法：
额，其实这才是最简单的方法，通过反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;  
    }  
};