Problem:

Reverse digits of an integer.

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!

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.

Solution:

To check for overflow/underflow, we could check if ret > 214748364 or ret < –214748364 before multiplying by 10. On the other hand, we do not need to check if ret == 214748364, why?

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题目大意：

给一个整数，要求将整数逆序得到另外一个整数，如果逆序后的整数越界，返回0,

解题思路：

直接用res=res*10+x/10;x/=10;来解决，注意判断越界，越界判断方法为转化为double类型然后判断结果是否超出INT_MAX（2147483647）

Java源代码（用时228ms）：

public class Solution {
    public int reverse(int x) {
        int flag=x>0?1:-1,res=0;
        x=x>0?x:-x;
        while(x>0){
            if(res*10.0 + x%10 > 2147483647)return 0;
            res = res*10+x%10;
            x/=10;
        }
        return res*flag;
    }
}

C语言源代码（用时10ms）：

int reverse(int x) {
    int flag=x>0?1:-1,res=0;
    x=x>0?x:-x;
    while(x>0){
        if((2147483647.0-x%10)/10<res)return 0;
        res=res*10+x%10;
        x=x/10;
    }
    return res*flag;
}

C++源代码（用时13ms）：

class Solution {
public:
    int reverse(int x) {
        int flag=x>0?1:-1,res=0;
        x=x>0?x:-x;
        while(x>0){
            if(res*10.0+x%10 > 2147483647)return 0;
            res = res*10+x%10;
            x/=10;
        }
        return res*flag;
    }
};

Python源代码（用时72ms）：

class Solution:
    # @param {integer} x
    # @return {integer}
    def reverse(self, x):
        flag=1 if x>0 else -1
        x=x if x>0 else -x
        res=0
        while x>0:
            if res*10.0 + x%10 > 2147483647:return 0
            res = res*10+x%10
            x/=10
        return res*flag