Reverse Integer Problem & Solution
Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range, then return 0.
See the reverse integer problem on LeetCode.
C++ Solution
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
static const int _=[](){std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();
class Solution {
public:
int reverse(int x) {
if (x == INT_MIN) {
return 0;
}
// abs(INT_MIN) == INT_MIN because of overflow, thus considering the edge case above.
int y = 0;
int tmp = abs(x);
while (tmp != 0) {
// Tests for overflow at each step.
// If you're going to do `y * 10` and it may overflow, then you need to test for `y > INT_MAX / 10`.
if (y > INT_MAX / 10 || y * 10 > INT_MAX - tmp % 10) {
return 0;
}
y = y * 10 + tmp % 10;
tmp /= 10;
}
if (x < 0) {
y *= -1;
}
return y;
}
};