LeetCode 7.Reverse digits of an integer.

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.

开始没有考虑到溢出的情况，有几个测试点没有通过。

对于32位int 整型范围为：-2^31~2^31-1

class Solution {
public:
    int reverse(int x) {
        if(x==0)
            return 0;
        int flag=x>0?1:0;
        x=x>0?x:-x;
        long long r=0;
        for(;x>0;x/=10){
            r=r*10+x%10;
        }
        if(r>2147483647||(flag==0&&r>2147483648)){
            return 0;
        }else{
        if(flag){
            return r;
        }
        else{
            return -r;
        }
        }
    }
};

同时看了下discussion里别人的解法，有几个地方可以改进：

（1）flag可以设置为-1和1，直接return flag*x，减少了条件判断

（2）可以用标准库文件<limits.h>里的INT_MIN和INT_MAX来代替MAX和Min取值

（3）如果要求不能使用long long 整型,则可以通过比较tmp和maximum/10，minimum/10的关系来确定

int reverse(int x) {    int minimum = INT_MIN, maximum = INT_MAX;    int res = 0, sign;    sign = (x<0)?-1:1;    if (x == minimum)        return 0;    x *= sign;  // here check x==minimum first    while (x) {        int tmp = sign*res;        if (tmp>maximum/10 || tmp<minimum/10 || (tmp==maximum/10 && x%10>maximum%10) || (tmp==minimum/10 && x%10<minimum%10))            return 0;        res = 10*res + x%10;        x /= 10;    }    return res*sign;}