【LeeCode】Reverse Integer 总结

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.

难点：如何判断int是否溢出？
总结：加上某值后再减去某值 与原值做对比 如果大小不变则没有溢出！

这是我的代码

class Solution {
public:
    int reverse(int x) {
        int ret=0;
        while(x!=0)
        {
            //如果大小改变则溢出，则返回0；
            if(ret!=ret*10/10)
            {
                return 0;
            }
            ret*=10;
            //同上
            if(ret!=ret+x%10-x%10)
            {
                return 0;
            }
            ret+=x%10;

            x=(x-x%10)/10;
        }
        return ret;
    }
};

这个是效率最高的代码：
还没有看，先放这里。

static int x = []() { 
    std::ios::sync_with_stdio(false); 
    cin.tie(NULL);  
    return 0; 
}();

class Solution {
public:
    int reverse(int x) {
        long answer = 0;
        while (x != 0) {
            answer = answer * 10 + x % 10;
            if (answer > INT_MAX || answer < INT_MIN) return 0;
            x /= 10;
        }
        return (int)answer;
    }
};