7. Reverse Integer

O(n) time O(1) space

class Solution:
    def reverse(self, x: int) -> int:
        res, MAX = 0, (1 << 31) - 1
        sign = -1 if x < 0 else 1
        x = abs(x)
        while x > 0:
            res = res * 10 + x % 10
            x //= 10
            if res > MAX: return 0
        return res * sign

这道题不需要保存负号，直接转即可

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

O(n) time O(n) space

class Solution {
public:
    int reverse(int x) {
        if (x == INT_MIN || x == 0) return 0;
        int sign = x < 0 ? -1 : 1;
        x = abs(x);
        string s = to_string(x);
        while (!s.empty() && s.back() == '0') s.pop_back();
        long res = 0;
        for (auto rit = s.rbegin(); rit != s.rend(); ++rit) {
            res = res * 10 + *rit - '0';
            if (res > INT_MAX) return 0;
        }
        return sign * res;
    }
};