LeetCode 7. Reverse Integer 一个整数倒叙输出

潜在问题：（1）随着求和可能精度会溢出int 范围，需要使用long 来辅助判断是否溢出，此时返回 0

               Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [?231,  231 ? 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

               （2）去前缀0

eg: reverse(1534236469); 会丢精度，如果不校验 所以WA了一次

int reverse(int x) {
    long sumLong = 0;
    int sum = 0;
    int num =  0;
    while (x!= 0) {    //支持正负数
        num = x % 10;  //末尾数字
        sum = sum * 10;//进位
        sum += num;
        x = x / 10;
        //校验精度
        sumLong = sumLong * 10;
        sumLong += num;
        if (sumLong != sum) {
            sum = 0;
            break;
        }
    }
    return sum;
}

LeetCode 7. Reverse Integer 一个整数倒叙输出

原文：https://www.cnblogs.com/someonelikeyou/p/9611139.html