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7.Reverse Integer

题目链接

要考虑到溢出问题,溢出问题可以用INT_MAX来比较,或者运算完之后再逆回去看时候和原数相等

int reverse(int x) {
    long val = 0;
	do 
	{
		val = val * 10 + x % 10;
		x /= 10;
	} while (x);
	
	return (val > INT_MAX || val < INT_MIN) ? 0 : val;
}