LEETCODE算法:7. Reverse Integer

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题目

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 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.

题解

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//转换字符串，翻转后在转成数
//可以实现，也是双百，但是用了long，不满足只能存32bits的条件
class Solution {
public:
    int reverse(int x) {
        string tmp=to_string(x);
        if(x>=0) std::reverse(tmp.begin(),tmp.end());
        else std::reverse(tmp.begin()+1,tmp.end());
        long ans=stol(tmp);
        if(ans>INT_MAX||ans<INT_MIN) return 0;
        return ans;
    }
};
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//核心问题在于要知道int区间的具体数值，当然可是当场打印看一下
class Solution {
public:
    int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int pop = x % 10;
            x /= 10;
            if (rev > INT_MAX/10 || (rev == INT_MAX / 10 && pop > 7)) return 0;
            if (rev < INT_MIN/10 || (rev == INT_MIN / 10 && pop < -8)) return 0;
            rev = rev * 10 + pop;
        }
        return rev;
    }
};