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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/someonelikeyou/p/9611139.html
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