LeetCode:Reverse Integer

Posted 2020-06-19 翎飞蝶舞

7. Reverse Integer

 

Total Accepted: 126996 Total Submissions: 538523 Difficulty: Easy

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.

 

 

    用各数位与int边界值比大小的方式，来确定是否越界。

class Solution {
public:
    int reverse(int x) {
        int neg=0;
        if (x<0)
        {
            neg=1;
            x=0-x;
        }
        int y=0;
        while(1)
        {
            int tmp1=x/10;
            int tmp2=x%10;
            if(neg!=1)
            {
                if(y>214748364)
                {
                    y=0;
                    break;
                }
                else if(y==214748364 && tmp2>7)
                {
                    y=0;
                    break;
                }
            }
            else
            {
                if(y>214748364)
                {
                    y=0;
                    break;
                }
                else if(y==214748364 && tmp2>8)
                {
                    y=0;
                    break;
                }
            }
            y=y*10+tmp2;
            x=tmp1;
            if(x==0)
                break;
        }
        if(neg==1)
            y=0-y;
        return y;
    }
};

        但是存在问题：若x=-2147483648，在第一步取反的时候就已经越界了。所以改用long long类型忽略越界的情况计算翻转值，最后再判断是否越界。

class Solution {
public:
    int reverse(int x) {
        int neg=0;
        long int y=0,tx=x;
        if (tx<0)
        {
            neg=1;
            tx=0-tx;
        }
      
        while(1)
        {
            int tmp1=tx/10;
            int tmp2=tx%10;
         
            y=y*10+tmp2;
            tx=tmp1;
            if(tx==0)
                break;
        }
        
        if(neg==1)
            y=0-y;
        if(y>INT_MAX || y<INT_MIN)
            return 0;
        return y;
    }
};