7. Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

int reverse(int x) {
    char strInt[12];
    sprintf(strInt,"%d",x);
    int n = 0;
    while(strInt[n] != '\0') n++;
    for(int i = 0,j = n - 1;i < j;i++,j--){
        if(strInt[i] == '-') i++;
        while(strInt[j] == '0'){
            strInt[j] = '\0';
            j--;
        }
        char tmp = strInt[i];
        strInt[i] = strInt[j];
        strInt[j] = tmp;
    }

    n = atoi(strInt);
    return n;
}

上面代码没有考虑内存溢出的问题，必须一位一位地转换。

int reverse(int x){
    int xr = 0,n = 1,p = x;
    while(p / 10){
        p /= 10;
        n *= 10;
    }
    p = 1;
    while(1){
        //判断是否内存溢出
        if(xr != (xr + p * (x / n)) % p) return 0;
        xr += p * (x / n);
        x %= n;
        p *= 10;
        n /= 10;
        if(x == 0) return xr;
    }
}

自己写的太丑了，附上别人的代码：

https://www.cnblogs.com/grandyang/p/5778281.html

class Solution {
public:
    /**
     * @param n the integer to be reversed
     * @return the reversed integer
     */
    int reverseInteger(int n) {
        int res = 0;
        while (n != 0) {
            int t = res * 10 + n % 10;
            if (t / 10 != res) return 0;
            res = t;
            n /= 10;
        }
        return res;
    }
};