【leetcode】整数反转

int reverse(int x){
    int arr[32] = {0};
    int bit = 0;
    long long Integer = 0;
    while(x)
    {
        arr[bit] = x % 10;
        x = x / 10;
        bit++;
    }
    int Loopbit = bit;
    bit --;
    for (int i=0; i<Loopbit; i++)
    {
        Integer += arr[i] * pow(10,bit--);
    }
    if(Integer < (int)-2147483648 | Integer > 2147483647)
        return 0;
    return Integer;
}
原文地址:https://www.cnblogs.com/ganxiang/p/13498053.html