LeetCode101: 7. Reverse Integer

• tags: Tricky,Stack,LeetCode101,Integer Overflow

The key is how to detect integer overflow without store to a larger size integer.

For this purpose, we could detect integer overflow before carry:

• The maximal of INT_MAX before carry is \(\frac{INT\_MAX}{10}\). We continue compare pop with the suffix of INT_MAX, 7, if maximal before carry is equal to \(\frac{INT\_MAX}{10}\).

• The minimal of INT_MIN before carry is \(\frac{INT\_MIN}{10}\) too. We continue compare pop with the suffix of INT_MIN, -8, if minimal before carry is equal to \(\frac{INT\_MIN}{10}\).

class Solution {
public:
	int reverse(int x) {
		// INT_MAX 2147483647
		// INT_MIN -2147483648
		int res = 0, pop = 0;
		while (abs(x) > 0) {
			pop = x % 10;
			x /= 10;
			// INT_MAX's suffix is 7
			if (res > INT_MAX / 10 || (res == INT_MAX / 10 && pop > 7)) {
				return 0;
			}
			// INT_MIN's suffix is -8
			if (res < INT_MIN / 10 || (res == INT_MIN / 10 && pop < -8)) {
				return 0;
			}
			res = res * 10 + pop;
		}
		return res;
	}
};